7 Parametric Equations and Polar Coordinates 参数方程与极坐标

7.4 Area and Arc Length in Polar Coordinates 7.4 极坐标中的面积与弧长

• 7.4.1 Apply the formula for area of a region in polar coordinates.

• 7.4.1 应用极坐标中区域面积的公式。

• 7.4.2 Determine the arc length of a polar curve.

• 7.4.2 确定极坐标曲线的弧长。

In the rectangular coordinate system, the definite integral provides a way to calculate the area under a curve. In particular, if we have a function $y = f(x)$ defined from $x = a$ to $x = b$ where $f(x) > 0$ on this interval, the area between the curve and the *x*-axis is given by $A = {\int_{a}^{b}{f(x)\ dx}}.$ This fact, along with the formula for evaluating this integral, is summarized in the Fundamental Theorem of Calculus. Similarly, the arc length of this curve is given by $L = {\int_{a}^{b}{\sqrt{1 + \left( {f^{\prime}(x)} \right)^{2}}dx}}.$ In this section, we study analogous formulas for area and arc length in the polar coordinate system.

Areas of Regions Bounded by Polar Curves 被极坐标曲线所围区域的面积

We have studied the formulas for area under a curve defined in rectangular coordinates and parametrically defined curves. Now we turn our attention to deriving a formula for the area of a region bounded by a polar curve. Recall that the proof of the Fundamental Theorem of Calculus used the concept of a Riemann sum to approximate the area under a curve by using rectangles. For polar curves we use the Riemann sum again, but the rectangles are replaced by sectors of a circle.

Consider a curve defined by the function $r = f(\theta),$ where $\alpha \leq \theta \leq \beta.$ Our first step is to partition the interval $\lbrack\alpha,\beta\rbrack$ into *n* equal-width subintervals. The width of each subinterval is given by the formula $\text{Δ}\theta = {{(\beta - \alpha)}\text{/}n},$ and the *i*th partition point $\theta_{i}$ is given by the formula $\theta_{i} = \alpha + i\text{Δ}\theta.$ Each partition point $\theta = \theta_{i}$ defines a line with slope $\text{tan}\theta_{i}$ passing through the pole as shown in the following graph.

The line segments are connected by arcs of constant radius. This defines sectors whose areas can be calculated by using a geometric formula. The area of each sector is then used to approximate the area between successive line segments. We then sum the areas of the sectors to approximate the total area. This approach gives a Riemann sum approximation for the total area. The formula for the area of a sector of a circle is illustrated in the following figure.

Recall that the area of a circle is $A = \pi r^{2}.$ When measuring angles in radians, 360 degrees is equal to $2\pi$ radians. Therefore a fraction of a circle can be measured by the central angle $\theta.$ The fraction of the circle is given by $\frac{\theta}{2\pi},$ so the area of the sector is this fraction multiplied by the total area:

$$A = \left( \frac{\theta}{2\pi} \right)\ \pi r^{2} = \frac{1}{2}\theta r^{2}.$$

Since the radius of a typical sector in Figure 7.39 is given by $r_{i} = f\left( \theta_{i} \right),$ the area of the *i*th sector is given by

$$A_{i} = \frac{1}{2}\left( {\text{Δ}\theta} \right)\left( {f\left( \theta_{i} \right)} \right)^{2}.$$

Therefore a Riemann sum that approximates the area is given by

$$A_{n} = {\sum\limits_{i = 1}^{n}A_{i}} \approx {\sum\limits_{i = 1}^{n}{\frac{1}{2}\left( {\text{Δ}\theta} \right)\left( {f\left( \theta_{i} \right)} \right)^{2}}}.$$

We take the limit as $n\rightarrow\infty$ to get the exact area:

$$A = \underset{n\rightarrow\infty}{\text{lim}}A_{n} = \frac{1}{2}{\int_{\alpha}^{\beta}{\left( {f(\theta)} \right)^{2}d\theta}}.$$

This gives the following theorem.

Area of a Region Bounded by a Polar Curve 由极坐标曲线所围区域的面积

Suppose $f$ is continuous and nonnegative on the interval $\alpha \leq \theta \leq \beta$ with $0 < \beta - \alpha \leq 2\pi.$ The area of the region bounded by the graph of $r = f(\theta)$ between the radial lines $\theta = \alpha$ and $\theta = \beta$ is

$$A = \frac{1}{2}{\int_{\alpha}^{\beta}{\left\lbrack {f(\theta)} \right\rbrack^{2}d\theta}} = \frac{1}{2}{\int_{\alpha}^{\beta}{r^{2}d\theta}}.$$ (7.9)

Finding an Area of a Polar Region 求极坐标区域的面积

Find the area of one petal of the rose defined by the equation $r = 3\ \text{sin}\left( {2\theta} \right).$

Solution 解答

The graph of $r = 3\ \text{sin}\left( {2\theta} \right)$ follows.

When $\theta = 0$ we have $r = 3\ \text{sin}\left( {2(0)} \right) = 0.$ The next value for which $r = 0$ is $\theta = \pi\text{/}2.$ This can be seen by solving the equation $3\ \text{sin}(2\theta) = 0$ for $\theta.$ Therefore the values $\theta = 0$ to $\theta = \pi\text{/}2$ trace out the first petal of the rose. To find the area inside this petal, use Equation 7.9 with $f(\theta) = 3\ \text{sin}\left( {2\theta} \right),$ $\alpha = 0,$ and $\beta = \pi\text{/}2\text{:}$

$$\begin{array}{cl} A & {= \frac{1}{2}{\int_{\alpha}^{\beta}{\left\lbrack {f(\theta)} \right\rbrack^{2}d\theta}}} \\ & {= \frac{1}{2}{\int_{0}^{\pi\text{/}2}{\left\lbrack {3\ \text{sin}\left( {2\theta} \right)} \right\rbrack^{2}d\theta}}} \\ & {= \frac{1}{2}{\int_{0}^{\pi\text{/}2}{9\ \text{sin}^{2}\left( {2\theta} \right)\ d\theta}}.} \end{array}$$

To evaluate this integral, use the formula $\text{sin}^{2}\alpha = \left( {1 - \text{cos}(2\alpha)} \right)\text{/}2$ with $\alpha = 2\theta\text{:}$

$$\begin{array}{cl} A & {= \frac{1}{2}{\int_{0}^{\pi\text{/}2}{9\ \text{sin}^{2}\left( {2\theta} \right)\ d\theta}}} \\ & {= \frac{9}{2}{\int_{0}^{\pi\text{/}2}{\frac{\left( {1 - \text{cos}\left( {4\theta} \right)} \right)}{2}d\theta}}} \\ & {= \frac{9}{4}\left( {\int_{0}^{\pi\text{/}2}{1 - \text{cos}\left( {4\theta} \right)\ d\theta}} \right)} \\ & {= \frac{9}{4}\left( \left. {\theta - \frac{\text{sin}\left( {4\theta} \right)}{4}} \right) \right._{0}^{\pi\text{/}2}} \\ & {= \frac{9}{4}\left( {\frac{\pi}{2} - \frac{\text{sin}\ 2\pi}{4}} \right) - \frac{9}{4}\left( {0 - \frac{\text{sin}\ 4(0)}{4}} \right)} \\ & {= \frac{9\pi}{8}.} \end{array}$$

Find the area inside the cardioid defined by the equation $r = 1 - \text{cos}\ \theta.$

Example 7.16 involved finding the area inside one curve. We can also use Area of a Region Bounded by a Polar Curve to find the area between two polar curves. However, we often need to find the points of intersection of the curves and determine which function defines the outer curve or the inner curve between these two points.

Finding the Area between Two Polar Curves 求两条极坐标曲线之间的面积

Find the area outside the cardioid $r = 2 + 2\ \text{sin}\ \theta$ and inside the circle $r = 6\ \text{sin}\ \theta.$

Solution 解答

First draw a graph containing both curves as shown.

To determine the limits of integration, first find the points of intersection by setting the two functions equal to each other and solving for $\theta\text{:}$

$$\begin{array}{rll} {6\ \text{sin}\ \theta} & = & {2 + 2\ \text{sin}\ \theta} \\ {4\ \text{sin}\ \theta} & = & 2 \\ {\text{sin}\ \theta} & = & {\frac{1}{2}.} \end{array}$$

This gives the solutions $\theta = \frac{\pi}{6}$ and $\theta = \frac{5\pi}{6},$ which are the limits of integration. The circle $r = 3\ \text{sin}\ \theta$ is the red graph, which is the outer function, and the cardioid $r = 2 + 2\ \text{sin}\ \theta$ is the blue graph, which is the inner function. To calculate the area between the curves, start with the area inside the circle between $\theta = \frac{\pi}{6}$ and $\theta = \frac{5\pi}{6},$ then subtract the area inside the cardioid between $\theta = \frac{\pi}{6}$ and $\theta = \frac{5\pi}{6}\text{:}$

$$\begin{array}{cl} A & {= \text{circle} - \text{cardioid}} \\ & {= \frac{1}{2}{\int_{\pi\text{/}6}^{{5\pi}\text{/}6}{\left\lbrack {6\ \text{sin}\ \theta} \right\rbrack^{2}d\theta}} - \frac{1}{2}{\int_{\pi\text{/}6}^{{5\pi}\text{/}6}{\left\lbrack {2 + 2\ \text{sin}\ \theta} \right\rbrack^{2}d\theta}}} \\ & {= \frac{1}{2}{\int_{\pi\text{/}6}^{{5\pi}\text{/}6}{36\ \text{sin}^{2}\theta\ d\theta}} - \frac{1}{2}{\int_{\pi\text{/}6}^{{5\pi}\text{/}6}{\left\lbrack 4 + 8\ \text{sin}\ \theta + 4\ \text{sin}^{2}\theta \right\rbrack\ d\theta}}} \\ & {= 18{\int_{\pi\text{/}6}^{{5\pi}\text{/}6}{\frac{1 - \text{cos}\left( {2\theta} \right)}{2}d\theta}} - 2{\int_{\pi\text{/}6}^{{5\pi}\text{/}6}{\left\lbrack 1 + 2\ \text{sin}\ \theta + \frac{1 - \text{cos}\left( {2\theta} \right)}{2} \right\rbrack d\theta}}} \\ & {= 9\left\lbrack {\theta - \frac{\text{sin}\left( {2\theta} \right)}{2}} \right\rbrack_{\pi\text{/}6}^{5\pi\text{/}6} - 2\left\lbrack {\frac{3\theta}{2} - 2\ \text{cos}\ \theta - \frac{\text{sin}\left( {2\theta} \right)}{4}} \right\rbrack_{\pi\text{/}6}^{5\pi\text{/}6}} \\ & {= 9\left( {\frac{5\pi}{6} - \frac{\text{sin}\ 2(5\pi\text{/}6)}{2}} \right) - 9\left( {\frac{\pi}{6} - \frac{\text{sin}\ 2(\pi\text{/}6)}{2}} \right)} \\ & {\mspace{7mu}\text{−}\left( {3\left( \frac{5\pi}{6} \right) - 4\ \text{cos}\ \frac{5\pi}{6} - \frac{\text{sin}\ 2(5\pi\text{/}6)}{2}} \right) + \left( {3\left( \frac{\pi}{6} \right) - 4\ \text{cos}\ \frac{\pi}{6} - \frac{\text{sin}\ 2(\pi\text{/}6)}{2}} \right)} \\ & {= 4\pi.} \end{array}$$

Find the area inside the circle $r = 4\ \text{cos}\ \theta$ and outside the circle $r = 2.$

In Example 7.17 we found the area inside the circle and outside the cardioid by first finding their intersection points. Notice that solving the equation directly for $\theta$ yielded two solutions: $\theta = \frac{\pi}{6}$ and $\theta = \frac{5\pi}{6}.$ However, in the graph there are three intersection points. The third intersection point is the origin. The reason why this point did not show up as a solution is because the origin is on both graphs but for different values of $\theta.$ For example, for the cardioid we get

$$\begin{array}{rll} {2 + 2\ \text{sin}\ \theta} & = & 0 \\ {\text{sin}\ \theta} & = & {-1,} \end{array}$$

so the values for $\theta$ that solve this equation are $\theta = \frac{3\pi}{2} + 2n\pi,$ where *n* is any integer. For the circle we get

$$6\ \text{sin}\ \theta = 0.$$

The solutions to this equation are of the form $\theta = n\pi$ for any integer value of *n.* These two solution sets have no points in common. Regardless of this fact, the curves intersect at the origin. This case must always be taken into consideration.

Arc Length in Polar Curves 极坐标曲线的弧长

Here we derive a formula for the arc length of a curve defined in polar coordinates.

In rectangular coordinates, the arc length of a parameterized curve $\left( {x(t),y(t)} \right)$ for $a \leq t \leq b$ is given by

$$L = {\int_{a}^{b}{\sqrt{\left( \frac{dx}{dt} \right)^{2} + \left( \frac{dy}{dt} \right)^{2}}dt}}.$$

In polar coordinates we define the curve by the equation $r = f(\theta),$ where $\alpha \leq \theta \leq \beta.$ In order to adapt the arc length formula for a polar curve, we use the equations

$$x = r\ \text{cos}\ \theta = f(\theta)\ \text{cos}\ \theta\ \text{and}\ y = r\ \text{sin}\ \theta = f(\theta)\ \text{sin}\ \theta,$$

and we replace the parameter *t* by $\theta.$ Then

$$\begin{array}{l} {\frac{dx}{d\theta} = f^{\prime}(\theta)\ \text{cos}\ \theta - f(\theta)\ \text{sin}\ \theta} \\ {\frac{dy}{d\theta} = f^{\prime}(\theta)\ \text{sin}\ \theta + f(\theta)\ \text{cos}\ \theta.} \end{array}$$

We replace $dt$ by $d\theta,$ and the lower and upper limits of integration are $\alpha$ and $\beta,$ respectively. Then the arc length formula becomes

$$\begin{array}{cl} L & {= {\int_{a}^{b}{\sqrt{\left( \frac{dx}{dt} \right)^{2} + \left( \frac{dy}{dt} \right)^{2}}dt}}} \\ & {= {\int_{\alpha}^{\beta}{\sqrt{\left( \frac{dx}{d\theta} \right)^{2} + \left( \frac{dy}{d\theta} \right)^{2}}d\theta}}} \\ & {= {\int_{\alpha}^{\beta}{\sqrt{\left( {f^{\prime}(\theta)\ \text{cos}\ \theta - f(\theta)\ \text{sin}\ \theta} \right)^{2} + \left( {f^{\prime}(\theta)\ \text{sin}\ \theta + f(\theta)\ \text{cos}\ \theta} \right)^{2}}d\theta}}} \\ & {= {\int_{\alpha}^{\beta}{\sqrt{\left( {f^{\prime}(\theta)} \right)^{2}\left( {\text{cos}^{2}\theta + \text{sin}^{2}\theta} \right) + \left( {f(\theta)} \right)^{2}\left( {\text{cos}^{2}\theta + \text{sin}^{2}\theta} \right)}d\theta}}} \\ & {= {\int_{\alpha}^{\beta}{\sqrt{\left( {f^{\prime}(\theta)} \right)^{2} + \left( {f(\theta)} \right)^{2}}d\theta}}} \\ & {= {\int_{\alpha}^{\beta}{\sqrt{r^{2} + \left( \frac{dr}{d\theta} \right)^{2}}d\theta}}.} \end{array}$$

This gives us the following theorem.

Arc Length of a Curve Defined by a Polar Function 由极坐标函数定义的曲线的弧长

Let $f$ be a function whose derivative is continuous on an interval $\alpha \leq \theta \leq \beta.$ The length of the graph of $r = f(\theta)$ from $\theta = \alpha$ to $\theta = \beta$ is

$$L = {\int_{\alpha}^{\beta}{\sqrt{\left\lbrack {f(\theta)} \right\rbrack^{2} + \left\lbrack {f^{\prime}(\theta)} \right\rbrack^{2}}d\theta = {\int_{\alpha}^{\beta}{\sqrt{r^{2} + \left( \frac{dr}{d\theta} \right)^{2}}d\theta.}}}}$$ (7.10)

Finding the Arc Length of a Polar Curve 求极坐标曲线的弧长

Find the arc length of the cardioid $r = 2 + 2\mspace{2mu}\text{cos}\mspace{2mu}\theta.$

Solution 解答

When $\theta = 0,r = 2 + 2\mspace{2mu}\text{cos}\mspace{2mu} 0 = 4.$ Furthermore, as $\theta$ goes from $0$ to $2\textit{π}\text{,}$ the cardioid is traced out exactly once. Therefore these are the limits of integration. Using $f{(\theta)} = 2 + 2\mspace{2mu}\text{cos}\mspace{2mu}\theta,$ $\alpha = 0,$ and $\beta = 2\textit{π}\text{,}$ Equation 7.10 becomes

$$\begin{array}{cl} L & {= {\int_{\alpha}^{\beta}\sqrt{\left\lbrack {f(\theta)} \right\rbrack^{2} + \left\lbrack {f^{\prime}(\theta)} \right\rbrack^{2}}}\ d\theta} \\ & {= {\int_{0}^{2\pi}\sqrt{\left\lbrack {2 + 2\mspace{2mu}\text{cos}\mspace{2mu}\theta} \right\rbrack^{2} + \left\lbrack {- 2\mspace{2mu}\text{sin}\mspace{2mu}\theta} \right\rbrack^{2}}}\ d\theta} \\ & {= {\int_{0}^{2\pi}\sqrt{4 + 8\mspace{2mu}\text{cos}\mspace{2mu}\theta + 4\mspace{2mu}\text{cos}^{2}\mspace{2mu}\theta + 4\mspace{2mu}\text{sin}^{2}\mspace{2mu}\theta}}\ d\theta} \\ & {= {\int_{0}^{2\pi}\sqrt{4 + 8\mspace{2mu}\text{cos}\mspace{2mu}\theta + 4\left( {\text{cos}^{2}\mspace{2mu}\theta + \text{sin}^{2}\mspace{2mu}\theta} \right)}}\ d\theta} \\ & {= {\int_{0}^{2\pi}\sqrt{8 + 8\mspace{2mu}\text{cos}\mspace{2mu}\theta}}\ d\theta} \\ & {= 2{\int_{0}^{2\pi}\sqrt{2 + 2\mspace{2mu}\text{cos}\mspace{2mu}\theta}}\ d\theta.} \end{array}$$

Next, using the identity $\text{cos}\left( {2\alpha} \right) = 2\mspace{2mu}\text{cos}^{2}\alpha - 1,$ add 1 to both sides and multiply by 2. This gives $2 + 2\mspace{2mu}\text{cos}\left( {2\alpha} \right) = 4\mspace{2mu}\text{cos}^{2}\alpha.$ Substituting $\alpha = {\theta\text{/}2}$ gives $2 + 2\mspace{2mu}\text{cos}\mspace{2mu}\theta = 4\mspace{2mu}\text{cos}^{2}({\theta\text{/}{2),}}$ so the integral becomes

$$\begin{array}{cl} L & {= 2{\int_{0}^{2\pi}\sqrt{2 + 2\ \text{cos}\ \theta}}d\theta} \\ & {= 2{\int_{0}^{2\pi}{\sqrt{4\ \text{cos}^{2}\left( \frac{\theta}{2} \right)}d\theta}}} \\ & {= 2{\int_{0}^{2\pi}{\left. 2 \middle| {\text{cos}\left( \frac{\theta}{2} \right)} \right|d\theta}}.} \end{array}$$

The absolute value is necessary because the cosine is negative for some values in its domain. To resolve this issue, change the limits from $0$ to $\pi$ and double the answer. This strategy works because cosine is positive between $0$ and $\frac{\pi}{2}.$ Thus,

$$\begin{array}{cl} L & {= 4{\int_{0}^{2\pi}{\left| {\text{cos}\left( \frac{\theta}{2} \right)} \right|d\theta}}} \\ & {= 8{\int_{0}^{\pi}{\text{cos}\left( \frac{\theta}{2} \right)\ d\theta}}} \\ & {= 8\left. \left( {2\ \text{sin}\left( \frac{\theta}{2} \right)} \right. \right)_{0}^{\pi}} \\ & {= 16.} \end{array}$$

Find the total arc length of $r = 3\ \text{sin}\ \theta.$

Section 7.4 Exercises 7.4 节习题

For the following exercises, determine a definite integral that represents the area.

188\.

Region enclosed by $r = 4$

189.

Region enclosed by $r = 3\ \text{sin}\ \theta$

190\.

Region in the first quadrant within the cardioid $r = 1 + \text{sin}\ \theta$

191.

Region enclosed by one petal of $r = 8\ \text{sin}(2\theta)$

192\.

Region enclosed by one petal of $r = \text{cos}(3\theta)$

193.

Region below the polar axis and enclosed by $r = 1 - \text{sin}\ \theta$

194\.

Region in the first quadrant enclosed by $r = 2 - \text{cos}\ \theta$

195.

Region enclosed by the inner loop of $r = 1 - 2\ \text{cos}\ \theta$

196\.

Region enclosed by the inner loop of $r = 3 - 4\ \text{cos}\ \theta$

197.

Region enclosed by $r = 1 - 2\ \text{cos}\ \theta$ and outside the inner loop

198\.

Region common to $r = 3\ \text{sin}\ \theta\ \text{and}\ r = 2 - \text{sin}\ \theta$

199.

Region common to $r = 2\ \text{and}\ r = 4\ \text{cos}\ \theta$

200\.

Region common to $r = 3\ \text{cos}\ \theta\ \text{and}\ r = 3\ \text{sin}\ \theta$

For the following exercises, find the area of the described region.

201.

Enclosed by $r = 6\ \text{sin}\ \theta$

202\.

Above the polar axis enclosed by $r = 2 + \text{sin}\ \theta$

203.

Below the polar axis and enclosed by $r = 2 - \text{cos}\ \theta$

204\.

Enclosed by one petal of $r = 4\ \text{cos}\left( {3\theta} \right)$

205.

Enclosed by one petal of $r = 3\ \text{cos}\left( {2\theta} \right)$

206\.

Enclosed by $r = 1 + \text{sin}\ \theta$

207.

Enclosed by the inner loop of $r = 3 + 6\ \text{cos}\ \theta$

208\.

Enclosed by $r = 2 + 4\ \text{cos}\ \theta$ and outside the inner loop

209.

Common interior of $r = 4\ \text{sin}\left( {2\theta} \right)\ \text{and}\ r = 2$

210\.

Common interior of $r = 3 - 2\ \text{sin}\ \theta\ \text{and}\ r = -3 + 2\ \text{sin}\ \theta$

211.

Common interior of $r = 6\ \text{sin}\ \theta\ \text{and}\ r = 3$

212\.

Inside $r = 1 + \text{cos}\ \theta$ and outside $r = \text{cos}\ \theta$

213.

Common interior of $r = 2 + 2\ \text{cos}\ \theta\ \text{and}\ r = 2\ \text{sin}\ \theta$

For the following exercises, find a definite integral that represents the arc length.

214\.

$r = 4\ \text{cos}\ \theta\ \text{on the interval}\ 0 \leq \theta \leq \frac{\pi}{2}$

215.

$r = 1 + \text{sin}\ \theta$ on the interval $0 \leq \theta \leq 2\pi$

216\.

$r = 2\ \text{sec}\ \theta\ \text{on the interval}\ 0 \leq \theta \leq \frac{\pi}{3}$

217.

$r = e^{\theta}\text{on the interval}\ 0 \leq \theta \leq 1$

For the following exercises, find the length of the curve over the given interval.

218\.

$r = 6\ \text{on the interval}\ 0 \leq \theta \leq \frac{\pi}{2}$

219.

$r = e^{3\theta}\text{on the interval}\ 0 \leq \theta \leq 2$

220\.

$r = 6\ \text{cos}\ \theta\ \text{on the interval}\ 0 \leq \theta \leq \frac{\pi}{2}$

221.

$r = 8 + 8\ \text{cos}\ \theta\ \text{on the interval}\ 0 \leq \theta \leq \pi$

222\.

$r = 1 - \text{sin}\ \theta\ \text{on the interval}\ 0 \leq \theta \leq 2\pi$

For the following exercises, use the integration capabilities of a calculator to approximate the length of the curve.

223.

\[T\] $r = 3\theta\ \text{on the interval}\ 0 \leq \theta \leq \frac{\pi}{2}$

224\.

\[T\] $r = \frac{2}{\theta}\ \text{on the interval}\ \pi \leq \theta \leq 2\pi$

225.

\[T\] $r = \text{sin}^{2}\left( \frac{\theta}{2} \right)\ \text{on the interval}\ 0 \leq \theta \leq \pi$

226\.

\[T\] $r = 2\theta^{2}\ \text{on the interval}\ 0 \leq \theta \leq \pi$

227.

\[T\] $r = \text{sin}\left( {3\ \text{cos}\ \theta} \right)\ \text{on the interval}\ 0 \leq \theta \leq \pi$

For the following exercises, use the familiar formula from geometry to find the area of the region described and then confirm by using the definite integral.

228\.

$r = 3\ \text{sin}\ \theta\ \text{on the interval}\ 0 \leq \theta \leq \pi$

229.

$r = \text{sin}\ \theta + \text{cos}\ \theta\ \text{on the interval}\ 0 \leq \theta \leq \pi$

230\.

$r = 6\ \text{sin}\ \theta + 8\ \text{cos}\ \theta\ \text{on the interval}\ 0 \leq \theta \leq \pi$

For the following exercises, use the familiar formula from geometry to find the length of the curve and then confirm using the definite integral.

231.

$r = 3\ \text{sin}\ \theta\ \text{on the interval}\ 0 \leq \theta \leq \pi$

232\.

$r = \text{sin}\ \theta + \text{cos}\ \theta\ \text{on the interval}\ 0 \leq \theta \leq \pi$

233.

$r = 6\ \text{sin}\ \theta + 8\ \text{cos}\ \theta\ \text{on the interval}\ 0 \leq \theta \leq \pi$

234\.

Verify that if $y = r\ \text{sin}\ \theta = f(\theta)\text{sin}\ \theta$ then $\frac{dy}{d\theta} = f\prime(\theta)\text{sin}\ \theta + f(\theta)\text{cos}\ \theta.$

For the following exercises, find the slope of a tangent line to a polar curve $r = f(\theta).$ Let $x = r\ \text{cos}\ \theta = f(\theta)\text{cos}\ \theta$ and $y = r\ \text{sin}\ \theta = f(\theta)\text{sin}\ \theta,$ so the polar equation $r = f(\theta)$ is now written in parametric form.

235.

Use the definition of the derivative $\frac{dy}{dx} = \frac{{dy}\text{/}{d\theta}}{{dx}\text{/}{d\theta}}$ and the product rule to derive the derivative of a polar equation.

236\.

$r = 1 - \text{sin}\ \theta;$ $\left( {\frac{1}{2},\frac{\pi}{6}} \right)$

237.

$r = 4\ \text{cos}\ \theta;$ $\left( {2,\frac{\pi}{3}} \right)$

238\.

$r = 8\ \text{sin}\ \theta;$ $\left( {4,\frac{5\pi}{6}} \right)$

239.

$r = 4 + \text{sin}\ \theta;$ $\left( {3,\frac{3\pi}{2}} \right)$

240\.

$r = 6 + 3\ \text{cos}\ \theta;$ $\left( {3,\pi} \right)$

241.

$r = 4\ \text{cos}\left( {2\theta} \right);$ tips of the leaves

242\.

$r = 2\ \text{sin}\left( {3\theta} \right);$ tips of the leaves

243.

$r = 2\theta;$ $\left( {\frac{\pi}{2},\frac{\pi}{4}} \right)$

244\.

Find the points on the interval $\text{−}\pi \leq \theta \leq \pi$ at which the cardioid $r = 1 - \text{cos}\ \theta$ has a vertical or horizontal tangent line.

245.

For the cardioid $r = 1 + \text{sin}\ \theta,$ find the slope of the tangent line when $\theta = \frac{\pi}{3}.$

For the following exercises, find the slope of the tangent line to the given polar curve at the point given by the value of $\theta.$

246\.

$r = 3\ \text{cos}\ \theta,\theta = \frac{\pi}{3}$

247.

$r = \theta,$ $\theta = \frac{\pi}{2}$

248\.

$r = \text{ln}\ \theta,$ $\theta = e$

249.

\[T\] Use technology: $r = 2 + 4\ \text{cos}\ \theta$ at $\theta = \frac{\pi}{6}$

For the following exercises, find the points at which the following polar curves have a horizontal or vertical tangent line.

250\.

$r = 4\ \text{cos}\ \theta$

251.

$r^{2} = 4\ \text{cos}\left( {2\theta} \right)$

252\.

$r = 2\ \text{sin}(2\theta)$

253.

The cardioid $r = 1 + \text{sin}\ \theta$

254\.

Show that the curve $r = \text{sin}\ \theta\ \text{tan}\ \theta$ (called a *cissoid of Diocles*) has the line $x = 1$ as a vertical asymptote.

7.5 Conic Sections 7.5 圆锥曲线

• 7.5.1 Identify the equation of a parabola in standard form with given focus and directrix.
• 7.5.2 Identify the equation of an ellipse in standard form with given foci.
• 7.5.3 Identify the equation of a hyperbola in standard form with given foci.
• 7.5.4 Recognize a parabola, ellipse, or hyperbola from its eccentricity value.
• 7.5.5 Write the polar equation of a conic section with eccentricity $e$.
• 7.5.6 Identify when a general equation of degree two is a parabola, ellipse, or hyperbola.

• 7.5.1 根据给定焦点和准线，识别抛物线的标准方程。
• 7.5.2 根据给定焦点，识别椭圆的标准方程。
• 7.5.3 根据给定焦点，识别双曲线的标准方程。
• 7.5.4 根据离心率的值识别抛物线、椭圆或双曲线。
• 7.5.5 写出离心率为 $e$ 的圆锥曲线的极坐标方程。
• 7.5.6 判断二次一般方程何时为抛物线、椭圆或双曲线。

Conic sections have been studied since the time of the ancient Greeks, and were considered to be an important mathematical concept. As early as 320 BCE, such Greek mathematicians as Menaechmus, Appollonius, and Archimedes were fascinated by these curves. Appollonius wrote an entire eight-volume treatise on conic sections in which he was, for example, able to derive a specific method for identifying a conic section through the use of geometry. Since then, important applications of conic sections have arisen (for example, in astronomy), and the properties of conic sections are used in radio telescopes, satellite dish receivers, and even architecture. In this section we discuss the three basic conic sections, some of their properties, and their equations.

Conic sections get their name because they can be generated by intersecting a plane with a cone. A cone has two identically shaped parts called nappes. One nappe is what most people mean by "cone," having the shape of a party hat. A right circular cone can be generated by revolving a line passing through the origin around the *y*-axis as shown.

Conic sections are generated by the intersection of a plane with a cone (Figure 7.44). If the plane intersects both nappes, then the conic section is a hyperbola. If the plane is parallel to the generating line, the conic section is a parabola. If the plane is perpendicular to the axis of revolution, the conic section is a circle. If the plane intersects one nappe at an angle to the axis (other than $90\text{°}),$ then the conic section is an ellipse.

Parabolas 抛物线

A parabola is generated when a plane intersects a cone parallel to the generating line. In this case, the plane intersects only one of the nappes. A parabola can also be defined in terms of distances.

A parabola is the set of all points whose distance from a fixed point, called the focus, is equal to the distance from a fixed line, called the directrix. The point halfway between the focus and the directrix is called the vertex of the parabola.

A graph of a typical parabola appears in Figure 7.45. Using this diagram in conjunction with the distance formula, we can derive an equation for a parabola. Recall the distance formula: Given point *P* with coordinates $\left( {x_{1},y_{1}} \right)$ and point *Q* with coordinates $\left( {x_{2},{\ \text{y}}_{2}} \right),$ the distance between them is given by the formula

$$d\left( {P,Q} \right) = \sqrt{\left( {x_{2} - x_{1}} \right)^{2} + \left( {y_{2} - y_{1}} \right)^{2}}.$$

Then from the definition of a parabola and Figure 7.45, we get

$$\begin{array}{rll} {d\left( {F,P} \right)} & = & {d\left( {P,Q} \right)} \\ \sqrt{\left( {0 - x} \right)^{2} + \left( {p - y} \right)^{2}} & = & {\sqrt{\left( {x - x} \right)^{2} + \left( {\text{−}p - y} \right)^{2}}.} \end{array}$$

Squaring both sides and simplifying yields

$$\begin{array}{rll} {x^{2} + \left( {p - y} \right)^{2}} & = & {0^{2} + \left( {\text{−}p - y} \right)^{2}} \\ {x^{2} + p^{2} - 2py + y^{2}} & = & {p^{2} + 2py + y^{2}} \\ {x^{2} - 2py} & = & {2py} \\ x^{2} & = & {4py.} \end{array}$$

Now suppose we want to relocate the vertex. We use the variables $\left( {h,k} \right)$ to denote the coordinates of the vertex. Then if the focus is directly above the vertex, it has coordinates $\left( {h,k + p} \right)$ and the directrix has the equation $y = k - p.$ Going through the same derivation yields the formula $\left( {x - h} \right)^{2} = 4p\left( {y - k} \right).$ Solving this equation for *y* leads to the following theorem.

Equations for Parabolas 抛物线的方程

Given a parabola opening upward with vertex located at $\left( {h,k} \right)$ and focus located at $\left( {h,k + p} \right),$ where *p* is a constant, the equation for the parabola is given by

$$y = \frac{1}{4p}\left( {x - h} \right)^{2} + k.$$ (7.11)

This is the standard form of a parabola.

We can also study the cases when the parabola opens down or to the left or the right. The equation for each of these cases can also be written in standard form as shown in the following graphs.

In addition, the equation of a parabola can be written in the general form, though in this form the values of *h*, *k*, and *p* are not immediately recognizable. The general form of a parabola is written as

$$ax^{2} + bx + cy + d = 0\quad\text{or}\quad ay^{2} + bx + cy + d = 0.$$

The first equation represents a parabola that opens either up or down. The second equation represents a parabola that opens either to the left or to the right. To put the equation into standard form, use the method of completing the square.

Converting the Equation of a Parabola from General into Standard Form 将抛物线方程由一般式化为标准式

Put the equation $x^{2} - 4x - 8y + 12 = 0$ into standard form and graph the resulting parabola.

Solution 解答

Since *y* is not squared in this equation, we know that the parabola opens either upward or downward. Therefore we need to solve this equation for *y,* which will put the equation into standard form. To do that, first add $8y$ to both sides of the equation:

$$8y = x^{2} - 4x + 12.$$

The next step is to complete the square on the right-hand side. Start by grouping the first two terms on the right-hand side using parentheses:

$$8y = \left( {x^{2} - 4x} \right) + 12.$$

Next determine the constant that, when added inside the parentheses, makes the quantity inside the parentheses a perfect square trinomial. To do this, take half the coefficient of *x* and square it. This gives $\left( \frac{-4}{2} \right)^{2} = 4.$ Add 4 inside the parentheses and subtract 4 outside the parentheses, so the value of the equation is not changed:

$$8y = \left( {x^{2} - 4x + 4} \right) + 12 - 4.$$

Now combine like terms and factor the quantity inside the parentheses:

$$8y = \left( {x - 2} \right)^{2} + 8.$$

Finally, divide by 8:

$$y = \frac{1}{8}\left( {x - 2} \right)^{2} + 1.$$

This equation is now in standard form. Comparing this to Equation 7.11 gives $h = 2,$ $k = 1,$ and $p = 2.$ The parabola opens up, with vertex at $\left( {2,1} \right),$ focus at $\left( {2,3} \right),$ and directrix $y = -1.$ The graph of this parabola appears as follows.

Put the equation $2y^{2} - x + 12y + 16 = 0$ into standard form and graph the resulting parabola.

The axis of symmetry of a vertical (opening up or down) parabola is a vertical line passing through the vertex. The parabola has an interesting reflective property. Suppose we have a satellite dish with a parabolic cross section. If a beam of electromagnetic waves, such as light or radio waves, comes into the dish in a straight line from a satellite (parallel to the axis of symmetry), then the waves reflect off the dish and collect at the focus of the parabola as shown.

Consider a parabolic dish designed to collect signals from a satellite in space. The dish is aimed directly at the satellite, and a receiver is located at the focus of the parabola. Radio waves coming in from the satellite are reflected off the surface of the parabola to the receiver, which collects and decodes the digital signals. This allows a small receiver to gather signals from a wide angle of sky. Flashlights and headlights in a car work on the same principle, but in reverse: the source of the light (that is, the light bulb) is located at the focus and the reflecting surface on the parabolic mirror focuses the beam straight ahead. This allows a small light bulb to illuminate a wide angle of space in front of the flashlight or car.

Ellipses 椭圆

An ellipse can also be defined in terms of distances. In the case of an ellipse, there are two foci (plural of focus), and two directrices (plural of directrix). We look at the directrices in more detail later in this section.

An *ellipse* is the set of all points for which the sum of their distances from two fixed points (the foci) is constant.

A graph of a typical ellipse is shown in Figure 7.48. In this figure the foci are labeled as $F$ and $F^{\prime}.$ Both are the same fixed distance from the origin, and this distance is represented by the variable *c*. Therefore the coordinates of $F$ are $\left( {c,0} \right)$ and the coordinates of $F^{\prime}$ are $\left( {\text{−}c,0} \right).$ The points $P$ and $P^{\prime}$ are located at the ends of the major axis of the ellipse, and have coordinates $\left( {a,0} \right)$ and $\left( {\text{−}a,0} \right),$ respectively. The major axis is always the longest distance across the ellipse, and can be horizontal or vertical. Thus, the length of the major axis in this ellipse is 2*a.* Furthermore, $P$ and $P^{\prime}$ are called the vertices of the ellipse. The points $Q$ and $Q^{\prime}$ are located at the ends of the minor axis of the ellipse, and have coordinates $\left( {0,b} \right)$ and $\left( {0,\text{−}b} \right),$ respectively. The minor axis is the shortest distance across the ellipse. The minor axis is perpendicular to the major axis.

According to the definition of the ellipse, we can choose any point on the ellipse and the sum of the distances from this point to the two foci is constant. Suppose we choose the point *P.* Since the coordinates of point *P* are $\left( {a,0} \right),$ the sum of the distances is

$$d\left( {P,F} \right) + d\left( {P,F^{\prime}} \right) = \left( {a - c} \right) + \left( {a + c} \right) = 2a.$$

Therefore the sum of the distances from an arbitrary point *A* with coordinates $\left( {x,y} \right)$ is also equal to 2*a.* Using the distance formula, we get

$$\begin{array}{rll} {d\left( {A,F} \right) + d\left( {A,F^{\prime}} \right)} & = & {2a} \\ {\sqrt{\left( {x - c} \right)^{2} + y^{2}} + \sqrt{\left( {x + c} \right)^{2} + y^{2}}} & = & {2a.} \end{array}$$

Subtract the second radical from both sides and square both sides:

$$\begin{array}{rll} \sqrt{\left( {x - c} \right)^{2} + y^{2}} & = & {2a - \sqrt{\left( {x + c} \right)^{2} + y^{2}}} \\ {\left( {x - c} \right)^{2} + y^{2}} & = & {4a^{2} - 4a\sqrt{\left( {x + c} \right)^{2} + y^{2}} + \left( {x + c} \right)^{2} + y^{2}} \\ {x^{2} - 2cx + c^{2} + y^{2}} & = & {4a^{2} - 4a\sqrt{\left( {x + c} \right)^{2} + y^{2}} + x^{2} + 2cx + c^{2} + y^{2}} \\ {\text{−}2cx} & = & {4a^{2} - 4a\sqrt{\left( {x + c} \right)^{2} + y^{2}} + 2cx.} \end{array}$$

Now isolate the radical on the right-hand side and square again:

$$\begin{array}{rll} {- 2cx} & = & {4a^{2} - 4a\sqrt{\left( {x + c} \right)^{2} + y^{2}} + 2cx} \\ {4a\sqrt{\left( {x + c} \right)^{2} + y^{2}}} & = & {4a^{2} + 4cx} \\ \sqrt{\left( {x + c} \right)^{2} + y^{2}} & = & {a + \frac{cx}{a}} \\ {\left( {x + c} \right)^{2} + y^{2}} & = & {a^{2} + 2cx + \frac{c^{2}x^{2}}{a^{2}}} \\ {x^{2} + 2cx + c^{2} + y^{2}} & = & {a^{2} + 2cx + \frac{c^{2}x^{2}}{a^{2}}} \\ {x^{2} + c^{2} + y^{2}} & = & {a^{2} + \frac{c^{2}x^{2}}{a^{2}}.} \end{array}$$

Isolate the variables on the left-hand side of the equation and the constants on the right-hand side:

$$\begin{array}{rll} & & \\ {x^{2} - \frac{c^{2}x^{2}}{a^{2}} + y^{2}} & = & {a^{2} - c^{2}} \\ {\frac{\left( {a^{2} - c^{2}} \right)x^{2}}{a^{2}} + y^{2}} & = & {a^{2} - c^{2}.} \end{array}$$

Divide both sides by $a^{2} - c^{2}.$ This gives the equation

$$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{a^{2} - c^{2}} = 1.$$

If we refer back to Figure 7.48, then the length of each of the two green line segments is equal to *a*. This is true because the sum of the distances from the point *Q* to the foci $F\ \text{and}\ F^{\prime}$ is equal to 2*a*, and the lengths of these two line segments are equal. This line segment forms a right triangle with hypotenuse length *a* and leg lengths *b* and *c*. From the Pythagorean theorem, $c^{2} = b^{2} = a^{2}$ and $b^{2} + a^{2} = c^{2}.$ Therefore the equation of the ellipse becomes

$$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1.$$

Finally, if the center of the ellipse is moved from the origin to a point $\left( {h,k} \right),$ we have the following standard form of an ellipse.

Equation of an Ellipse in Standard Form 椭圆的标准方程

Consider the ellipse with center $\left( {h,k} \right),$ a horizontal major axis with length 2*a*, and a vertical minor axis with length 2*b*. Then the equation of this ellipse in standard form is

$$\frac{\left( {x - h} \right)^{2}}{a^{2}} + \frac{\left( {y - k} \right)^{2}}{b^{2}} = 1$$ (7.12)

and the foci are located at $\left( {h \pm c,k} \right),$ where $c^{2} = a^{2} - b^{2}.$ The equations of the directrices are $x = h \pm \frac{a^{2}}{c}.$

If the major axis is vertical, then the equation of the ellipse becomes

$$\frac{\left( {x - h} \right)^{2}}{b^{2}} + \frac{\left( {y - k} \right)^{2}}{a^{2}} = 1$$ (7.13)

and the foci are located at $\left( {h,k \pm c} \right),$ where $c^{2} = a^{2} - b^{2}.$ The equations of the directrices in this case are $y = k \pm \frac{a^{2}}{c}.$

If the major axis is horizontal, then the ellipse is called horizontal, and if the major axis is vertical, then the ellipse is called vertical. The equation of an ellipse is in general form if it is in the form $Ax^{2} + By^{2} + Cx + Dy + E = 0,$ where *A* and *B* are either both positive or both negative. To convert the equation from general to standard form, use the method of completing the square.

Finding the Standard Form of an Ellipse 求椭圆的标准方程

Put the equation $9x^{2} + 4y^{2} - 36x + 24y + 36 = 0$ into standard form and graph the resulting ellipse.

Solution 解答

First subtract 36 from both sides of the equation:

$$9x^{2} + 4y^{2} - 36x + 24y = -36.$$

Next group the *x* terms together and the *y* terms together, and factor out the common factor:

$$\begin{array}{rll} {\left( {9x^{2} - 36x} \right) + \left( {4y^{2} + 24y} \right)} & = & -36 \\ {9\left( {x^{2} - 4x} \right) + 4\left( {y^{2} + 6y} \right)} & = & -36. \end{array}$$

We need to determine the constant that, when added inside each set of parentheses, results in a perfect square. In the first set of parentheses, take half the coefficient of *x* and square it. This gives $\left( \frac{-4}{2} \right)^{2} = 4.$ In the second set of parentheses, take half the coefficient of *y* and square it. This gives $\left( \frac{6}{2} \right)^{2} = 9.$ Add these inside each pair of parentheses. Since the first set of parentheses has a 9 in front, we are actually adding 36 to the left-hand side. Similarly, we are adding 36 to the second set as well. Therefore the equation becomes

$$\begin{array}{l} \\ {9\left( {x^{2} - 4x + 4} \right) + 4\left( {y^{2} + 6y + 9} \right) = -36 + 36 + 36} \\ {9\left( {x^{2} - 4x + 4} \right) + 4\left( {y^{2} + 6y + 9} \right) = 36.} \end{array}$$

Now factor both sets of parentheses and divide by 36:

$$\begin{array}{rll} & & \\ {9\left( {x - 2} \right)^{2} + 4\left( {y + 3} \right)^{2}} & = & 36 \\ {\frac{9\left( {x - 2} \right)^{2}}{36} + \frac{4\left( {y + 3} \right)^{2}}{36}} & = & 1 \\ {\frac{\left( {x - 2} \right)^{2}}{4} + \frac{\left( {y + 3} \right)^{2}}{9}} & = & 1. \end{array}$$

The equation is now in standard form. Comparing this to Equation 7.14 gives $h = 2,$ $k = -3,$ $a = 3,$ and $b = 2.$ This is a vertical ellipse with center at $\left( {2,-3} \right),$ major axis 6, and minor axis 4. The graph of this ellipse appears as follows.

Put the equation $9x^{2} + 16y^{2} + 18x - 64y - 71 = 0$ into standard form and graph the resulting ellipse.

According to Kepler’s first law of planetary motion, the orbit of a planet around the Sun is an ellipse with the Sun at one of the foci as shown in Figure 7.50(a). Because Earth’s orbit is an ellipse, the distance from the Sun varies throughout the year. A commonly held misconception is that Earth is closer to the Sun in the summer. In fact, in summer for the northern hemisphere, Earth is farther from the Sun than during winter. The difference in season is caused by the tilt of Earth’s axis in the orbital plane. Comets that orbit the Sun, such as Halley’s Comet, also have elliptical orbits, as do moons orbiting the planets and satellites orbiting Earth.

Ellipses also have interesting reflective properties: A light ray emanating from one focus passes through the other focus after mirror reflection in the ellipse. The same thing occurs with a sound wave as well. The National Statuary Hall in the U.S. Capitol in Washington, DC, is a famous room in an elliptical shape as shown in Figure 7.50(b). This hall served as the meeting place for the U.S. House of Representatives for almost fifty years. The location of the two foci of this semi-elliptical room are clearly identified by marks on the floor, and even if the room is full of visitors, when two people stand on these spots and speak to each other, they can hear each other much more clearly than they can hear someone standing close by. Legend has it that John Quincy Adams had his desk located on one of the foci and was able to eavesdrop on everyone else in the House without ever needing to stand. Although this makes a good story, it is unlikely to be true, because the original ceiling produced so many echoes that the entire room had to be hung with carpets to dampen the noise. The ceiling was rebuilt in 1902 and only then did the now-famous whispering effect emerge. Another famous whispering gallery—the site of many marriage proposals—is in Grand Central Station in New York City.

Hyperbolas 双曲线

A hyperbola can also be defined in terms of distances. In the case of a hyperbola, there are two foci and two directrices. Hyperbolas also have two asymptotes.

A hyperbola is the set of all points where the difference between their distances from two fixed points (the foci) is constant.

A graph of a typical hyperbola appears as follows.

The derivation of the equation of a hyperbola in standard form is virtually identical to that of an ellipse. One slight hitch lies in the definition: The difference between two numbers is always positive. Let *P* be a point on the hyperbola with coordinates $\left( {x,y} \right).$ Then the definition of the hyperbola gives $\left| {d\left( {P,F_{1}} \right) - d\left( {P,F_{2}} \right)} \right| = \text{constant}.$ To simplify the derivation, assume that *P* is on the right branch of the hyperbola, so the absolute value bars drop. If it is on the left branch, then the subtraction is reversed. The vertex of the right branch has coordinates $\left( {a,0} \right),$ so

$$d\left( {P,F_{1}} \right) - d\left( {P,F_{2}} \right) = \left( {c + a} \right) - \left( {c - a} \right) = 2a.$$

This equation is therefore true for any point on the hyperbola. Returning to the coordinates $\left( {x,y} \right)$ for *P*:

$$\begin{array}{rll} {d\left( {P,F_{1}} \right) - d\left( {P,F_{2}} \right)} & = & {2a} \\ {\sqrt{\left( {x + c} \right)^{2} + y^{2}} - \sqrt{\left( {x - c} \right)^{2} + y^{2}}} & = & {2a.} \end{array}$$

Add the second radical from both sides and square both sides:

$$\begin{array}{rll} \sqrt{\left( {x - c} \right)^{2} + y^{2}} & = & {2a + \sqrt{\left( {x + c} \right)^{2} + y^{2}}} \\ {\left( {x - c} \right)^{2} + y^{2}} & = & {4a^{2} + 4a\sqrt{\left( {x + c} \right)^{2} + y^{2}} + \left( {x + c} \right)^{2} + y^{2}} \\ {x^{2} - 2cx + c^{2} + y^{2}} & = & {4a^{2} + 4a\sqrt{\left( {x + c} \right)^{2} + y^{2}} + x^{2} + 2cx + c^{2} + y^{2}} \\ {\text{−}2cx} & = & {4a^{2} + 4a\sqrt{\left( {x + c} \right)^{2} + y^{2}} + 2cx.} \end{array}$$

Now isolate the radical on the right-hand side and square again:

$$\begin{array}{rll} {- 2cx} & = & {4a^{2} + 4a\sqrt{\left( {x + c} \right)^{2} + y^{2}} + 2cx} \\ {4a\sqrt{\left( {x + c} \right)^{2} + y^{2}}} & = & {-4a^{2} - 4cx} \\ \sqrt{\left( {x + c} \right)^{2} + y^{2}} & = & {\text{−}a - \frac{cx}{a}} \\ {\left( {x + c} \right)^{2} + y^{2}} & = & {a^{2} + 2cx + \frac{c^{2}x^{2}}{a^{2}}} \\ {x^{2} + 2cx + c^{2} + y^{2}} & = & {a^{2} + 2cx + \frac{c^{2}x^{2}}{a^{2}}} \\ {x^{2} + c^{2} + y^{2}} & = & {a^{2} + \frac{c^{2}x^{2}}{a^{2}}.} \end{array}$$

Isolate the variables on the left-hand side of the equation and the constants on the right-hand side:

$$\begin{array}{rll} & & \\ {x^{2} - \frac{c^{2}x^{2}}{a^{2}} + y^{2}} & = & {a^{2} - c^{2}} \\ {\frac{\left( {a^{2} - c^{2}} \right)x^{2}}{a^{2}} + y^{2}} & = & {a^{2} - c^{2}.} \end{array}$$

Finally, divide both sides by $a^{2} - c^{2}.$ This gives the equation

$$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{a^{2} - c^{2}} = 1.$$

We now define *b* so that $b^{2} = c^{2} - a^{2}.$ This is possible because $c > a.$ Therefore the equation of the hyperbola becomes

$$\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1.$$

Finally, if the center of the hyperbola is moved from the origin to the point $\left( {h,k} \right),$ we have the following standard form of a hyperbola.

Equation of a Hyperbola in Standard Form 标准形式的双曲线方程

Consider the hyperbola with center $\left( {h,k} \right),$ a horizontal major axis, and a vertical minor axis. Then the equation of this hyperbola is

$$\frac{\left( {x - h} \right)^{2}}{a^{2}} - \frac{\left( {y - k} \right)^{2}}{b^{2}} = 1$$ (7.14)

and the foci are located at $\left( {h \pm c,k} \right),$ where $c^{2} = a^{2} + b^{2}.$ The equations of the asymptotes are given by $y = k \pm \frac{b}{a}\left( {x - h} \right).$ The equations of the directrices are

$$x = h \pm \frac{a^{2}}{\sqrt{a^{2} + b^{2}}} = h \pm \frac{a^{2}}{c}.$$

If the major axis is vertical, then the equation of the hyperbola becomes

$$\frac{\left( {y - k} \right)^{2}}{a^{2}} - \frac{\left( {x - h} \right)^{2}}{b^{2}} = 1$$ (7.15)

and the foci are located at $\left( {h,k \pm c} \right),$ where $c^{2} = a^{2} + b^{2}.$ The equations of the asymptotes are given by $y = k \pm \frac{a}{b}\left( {x - h} \right).$ The equations of the directrices are

$$y = k \pm \frac{a^{2}}{\sqrt{a^{2} + b^{2}}} = k \pm \frac{a^{2}}{c}.$$

If the major axis (transverse axis) is horizontal, then the hyperbola is called horizontal, and if the major axis is vertical then the hyperbola is called vertical. The equation of a hyperbola is in general form if it is in the form $Ax^{2} + By^{2} + Cx + Dy + E = 0,$ where *A* and *B* have opposite signs. In order to convert the equation from general to standard form, use the method of completing the square.

Finding the Standard Form of a Hyperbola 求双曲线的标准形式

Put the equation $9x^{2} - 16y^{2} + 36x + 32y - 124 = 0$ into standard form and graph the resulting hyperbola. What are the equations of the asymptotes?

Solution

First add 124 to both sides of the equation:

$$9x^{2} - 16y^{2} + 36x + 32y = 124.$$

Next group the *x* terms together and the *y* terms together, then factor out the common factors:

$$\begin{array}{rll} {\left( {9x^{2} + 36x} \right) - \left( {16y^{2} - 32y} \right)} & = & 124 \\ {9\left( {x^{2} + 4x} \right) - 16\left( {y^{2} - 2y} \right)} & = & 124. \end{array}$$

We need to determine the constant that, when added inside each set of parentheses, results in a perfect square. In the first set of parentheses, take half the coefficient of *x* and square it. This gives $\left( \frac{4}{2} \right)^{2} = 4.$ In the second set of parentheses, take half the coefficient of *y* and square it. This gives $\left( \frac{-2}{2} \right)^{2} = 1.$ Add these inside each pair of parentheses. Since the first set of parentheses has a 9 in front, we are actually adding 36 to the left-hand side. Similarly, we are subtracting 16 from the second set of parentheses. Therefore the equation becomes

$$\begin{array}{l} \\ {9\left( {x^{2} + 4x + 4} \right) - 16\left( {y^{2} - 2y + 1} \right) = 124 + 36 - 16} \\ {9\left( {x^{2} + 4x + 4} \right) - 16\left( {y^{2} - 2y + 1} \right) = 144.} \end{array}$$

Next factor both sets of parentheses and divide by 144:

$$\begin{array}{rll} & & \\ {9\left( {x + 2} \right)^{2} - 16\left( {y - 1} \right)^{2}} & = & 144 \\ {\frac{9\left( {x + 2} \right)^{2}}{144} - \frac{16\left( {y - 1} \right)^{2}}{144}} & = & 1 \\ {\frac{\left( {x + 2} \right)^{2}}{16} - \frac{\left( {y - 1} \right)^{2}}{9}} & = & 1. \end{array}$$

The equation is now in standard form. Comparing this to Equation 7.15 gives $h = -2,$ $k = 1,$ $a = 4,$ and $b = 3.$ This is a horizontal hyperbola with center at $\left( {-2,1} \right)$ and asymptotes given by the equations $y = 1 \pm \frac{3}{4}\left( {x + 2} \right).$ The graph of this hyperbola appears in the following figure.

Put the equation $4y^{2} - 9x^{2} + 16y + 18x - 29 = 0$ into standard form and graph the resulting hyperbola. What are the equations of the asymptotes?

Hyperbolas also have interesting reflective properties. A ray directed toward one focus of a hyperbola is reflected by a hyperbolic mirror toward the other focus. This concept is illustrated in the following figure.

This property of the hyperbola has important applications. It is used in radio direction finding (since the difference in signals from two towers is constant along hyperbolas), and in the construction of mirrors inside telescopes (to reflect light coming from the parabolic mirror to the eyepiece). Another interesting fact about hyperbolas is that for a comet entering the solar system, if the speed is great enough to escape the Sun’s gravitational pull, then the path that the comet takes as it passes through the solar system is hyperbolic.

Eccentricity and Directrix 离心率与准线

An alternative way to describe a conic section involves the directrices, the foci, and a new property called eccentricity. We will see that the value of the eccentricity of a conic section can uniquely define that conic.

The eccentricity *e* of a conic section is defined to be the distance from any point on the conic section to its focus, divided by the perpendicular distance from that point to the nearest directrix. This value is constant for any conic section, and can define the conic section as well:

1. If $e = 1,$ the conic is a parabola.

2. If $e < 1,$ it is an ellipse.

3. If $e > 1,$ it is a hyperbola.

The eccentricity of a circle is zero. The directrix of a conic section is the line that, together with the point known as the focus, serves to define a conic section. Hyperbolas and noncircular ellipses have two foci and two associated directrices. Parabolas have one focus and one directrix.

The three conic sections with their directrices appear in the following figure.

Recall from the definition of a parabola that the distance from any point on the parabola to the focus is equal to the distance from that same point to the directrix. Therefore, by definition, the eccentricity of a parabola must be 1. The equations of the directrices of a horizontal ellipse are $x = \text{±}\frac{a^{2}}{c}.$ The right vertex of the ellipse is located at $\left( {a,0} \right)$ and the right focus is $\left( {c,0} \right).$ Therefore the distance from the vertex to the focus is $a - c$ and the distance from the vertex to the right directrix is $\frac{a^{2}}{c} - a.$ This gives the eccentricity as

$$e = \frac{a - c}{\frac{a^{2}}{c} - a} = \frac{c\left( {a - c} \right)}{a^{2} - ac} = \frac{c\left( {a - c} \right)}{a\left( {a - c} \right)} = \frac{c}{a}.$$

Since $c < a,$ this step proves that the eccentricity of an ellipse is less than 1. The directrices of a horizontal hyperbola are also located at $x = \text{±}\frac{a^{2}}{c},$ and a similar calculation shows that the eccentricity of a hyperbola is also $e = \frac{c}{a}.$ However in this case we have $c > a,$ so the eccentricity of a hyperbola is greater than 1.

Determining Eccentricity of a Conic Section 确定圆锥曲线的离心率

Determine the eccentricity of the ellipse described by the equation

$$\frac{\left( {x - 3} \right)^{2}}{16} + \frac{\left( {y + 2} \right)^{2}}{25} = 1.$$

Solution

From the equation we see that $a = 5$ and $b = 4.$ The value of *c* can be calculated using the equation $a^{2} = b^{2} + c^{2}$ for an ellipse. Substituting the values of *a* and *b* and solving for *c* gives $c = 3.$ Therefore the eccentricity of the ellipse is $e = \frac{c}{a} = \frac{3}{5} = 0.6.$

Determine the eccentricity of the hyperbola described by the equation

$$\frac{\left( {y - 3} \right)^{2}}{49} - \frac{\left( {x + 2} \right)^{2}}{25} = 1.$$

Polar Equations of Conic Sections 圆锥曲线的极坐标方程

Sometimes it is useful to write or identify the equation of a conic section in polar form. To do this, we need the concept of the focal parameter. The focal parameter of a conic section *p* is defined as the distance from a focus to the nearest directrix. The following table gives the focal parameters for the different types of conics, where *a* is the length of the semi-major axis (i.e., half the length of the major axis), *c* is the distance from the origin to the focus, and *e* is the eccentricity. In the case of a parabola, *a* represents the distance from the vertex to the focus.

Using the definitions of the focal parameter and eccentricity of the conic section, we can derive an equation for any conic section in polar coordinates. In particular, we assume that one of the foci of a given conic section lies at the pole. Then using the definition of the various conic sections in terms of distances, it is possible to prove the following theorem.

Polar Equation of Conic Sections 圆锥曲线的极坐标方程

The polar equation of a conic section with focal parameter *p* is given by

$$r = \frac{ep}{1 \pm e\ \text{cos}\ \theta}\ \text{or}\ r = \frac{ep}{1 \pm e\ \text{sin}\ \theta}.$$

In the equation on the left, the major axis of the conic section is horizontal, and in the equation on the right, the major axis is vertical. To work with a conic section written in polar form, first make the constant term in the denominator equal to 1. This can be done by dividing both the numerator and the denominator of the fraction by the constant that appears in front of the plus or minus in the denominator. Then the coefficient of the sine or cosine in the denominator is the eccentricity. This value identifies the conic. If cosine appears in the denominator, then the conic is horizontal. If sine appears, then the conic is vertical. If both appear then the axes are rotated. The center of the conic is not necessarily at the origin. The center is at the origin only if the conic is a circle (i.e., $e = 0).$

Graphing a Conic Section in Polar Coordinates 在极坐标中绘制圆锥曲线

Identify and create a graph of the conic section described by the equation

$$r = \frac{3}{1 + 2\ \text{cos}\ \theta}.$$

Solution

The constant term in the denominator is 1, so the eccentricity of the conic is 2. This is a hyperbola. The focal parameter *p* can be calculated by using the equation $ep = 3.$ Since $e = 2,$ this gives $p = \frac{3}{2}.$ The cosine function appears in the denominator, so the hyperbola is horizontal. Pick a few values for $\theta$ and create a table of values. Then we can graph the hyperbola (Figure 7.55).

Identify and create a graph of the conic section described by the equation

$$r = \frac{4}{1 - 0.8\ \text{sin}\ \theta}.$$

General Equations of Degree Two 二次一般方程

A general equation of degree two can be written in the form

$$Ax^{2} + Bxy + Cy^{2} + Dx + Ey + F = 0.$$

The graph of an equation of this form is a conic section. If $B \neq 0$ then the coordinate axes are rotated. To identify the conic section, we use the discriminant of the conic section $4AC - B^{2}.$ One of the following cases must be true:

1. $4AC - B^{2} > 0.$ If so, the graph is an ellipse.

2. $4AC - B^{2} = 0.$ If so, the graph is a parabola.

3. $4AC - B^{2} < 0.$ If so, the graph is a hyperbola.

The simplest example of a second-degree equation involving a cross term is $xy = 1.$ This equation can be solved for *y* to obtain $y = \frac{1}{x}.$ The graph of this function is called a *rectangular hyperbola* as shown.

The asymptotes of this hyperbola are the *x* and *y* coordinate axes. To determine the angle $\theta$ of rotation of the conic section, we use the formula $\text{cot}\ 2\theta = \frac{A - C}{B}.$ In this case $A = C = 0$ and $B = 1,$ so $\text{cot}\ 2\theta = {{(0 - 0)}\text{/}{1 = 0}}$ and $\theta = 45\text{°}.$ The method for graphing a conic section with rotated axes involves determining the coefficients of the conic in the rotated coordinate system. The new coefficients are labeled $A^{\prime},B^{\prime},C^{\prime},D^{\prime},E^{\prime},\ \text{and}\ F^{\prime},$ and are given by the formulas

$$\begin{array}{rll} A^{\prime} & = & {A\ \text{cos}^{2}\theta + B\ \text{cos}\ \theta\ \text{sin}\ \theta + C\ \text{sin}^{2}\theta} \\ B^{\prime} & = & 0 \\ C^{\prime} & = & {A\ \text{sin}^{2}\theta - B\ \text{sin}\ \theta\ \text{cos}\ \theta + C\ \text{cos}^{2}\theta} \\ D^{\prime} & = & {D\ \text{cos}\ \theta + E\ \text{sin}\ \theta} \\ E^{\prime} & = & {\text{−}D\ \text{sin}\ \theta + E\ \text{cos}\ \theta} \\ F^{\prime} & = & {F.} \end{array}$$

The procedure for graphing a rotated conic is the following:

1. Identify the conic section using the discriminant $4AC - B^{2}.$

2. Determine $\theta$ using the formula $\text{cot}\ 2\theta = \frac{A - C}{B}.$

3. Calculate $A^{\prime},B^{\prime},C^{\prime},D^{\prime},E^{\prime},\ \text{and}\ F^{\prime}.$

4. Rewrite the original equation using $A^{\prime},B^{\prime},C^{\prime},D^{\prime},E^{\prime},\ \text{and}\ F^{\prime}.$

5. Draw a graph using the rotated equation.

Identifying a Rotated Conic 识别旋转圆锥曲线

Identify the conic and calculate the angle of rotation of axes for the curve described by the equation

$$13x^{2} - 6\sqrt{3}xy + 7y^{2} - 256 = 0.$$

Solution 解答

In this equation, $A = 13,B = -6\sqrt{3},C = 7,D = 0,E = 0,$ and $F = -256.$ The discriminant of this equation is $4AC - B^{2} = 4(13)(7) - \left( {-6\sqrt{3}} \right)^{2} = 364 - 108 = 256.$ Therefore this conic is an ellipse. To calculate the angle of rotation of the axes, use $\text{cot}\ 2\theta = \frac{A - C}{B}.$ This gives

$$\begin{array}{cl} {\text{cot}\ 2\theta} & {= \frac{A - C}{B}} \\ & {= \frac{13 - 7}{-6\sqrt{3}}} \\ & {= - \frac{\sqrt{3}}{3}.} \end{array}$$

Therefore $2\theta = 120^{\text{o}}$ and $\theta = 60^{\text{o}},$ which is the angle of the rotation of the axes.

To determine the rotated coefficients, use the formulas given above:

$$\begin{array}{cll} A^{\prime} & = & {A\ \text{cos}^{2}\theta + B\ \text{cos}\ \theta\ \text{sin}\ \theta + C\ \text{sin}^{2}\theta} \\ & = & {13\text{cos}^{2}60 + \left( {-6\sqrt{3}} \right)\ \text{cos}\ 60\ \text{sin}\ 60 + 7\text{sin}^{2}60} \\ & = & {13\left( \frac{1}{2} \right)^{2} - 6\sqrt{3}\left( \frac{1}{2} \right)\left( \frac{\sqrt{3}}{2} \right) + 7\left( \frac{\sqrt{3}}{2} \right)^{2}} \\ & = & 4, \\ B^{\prime} & = & {0,} \\ C^{\prime} & = & {A\ \text{sin}^{2}\theta - B\ \text{sin}\ \theta\ \text{cos}\ \theta + C\ \text{cos}^{2}\theta} \\ & = & {13\text{sin}^{2}60 + \left( {-6\sqrt{3}} \right)\ \text{sin}\ 60\ \text{cos}\ 60 = 7\text{cos}^{2}60} \\ & = & {\left( \frac{\sqrt{3}}{2} \right)^{2} + 6\sqrt{3}\left( \frac{\sqrt{3}}{2} \right)\left( \frac{1}{2} \right) + 7\left( \frac{1}{2} \right)^{2}} \\ & = & 16, \\ D^{\prime} & = & {D\ \text{cos}\ \theta + E\ \text{sin}\ \theta} \\ & = & {(0)\ \text{cos}\ 60 + (0)\ \text{sin}\ 60} \\ & = & 0, \\ E^{\prime} & = & {\text{−}D\ \text{sin}\ \theta + E\ \text{cos}\ \theta} \\ & = & {\text{−}(0)\ \text{sin}\ 60 + (0)\ \text{cos}\ 60} \\ & = & 0, \\ F^{\prime} & = & F \\ & = & -256. \end{array}$$

The equation of the conic in the rotated coordinate system becomes

$$\begin{array}{rll} & & \\ {4\left( x^{\prime} \right)^{2} + 16\left( y^{\prime} \right)^{2}} & = & 256 \\ {\frac{\left( x^{\prime} \right)^{2}}{64} + \frac{\left( y^{\prime} \right)^{2}}{16}} & = & 1. \end{array}$$

A graph of this conic section appears as follows.

Identify the conic and calculate the angle of rotation of axes for the curve described by the equation

$$3x^{2} + 5xy - 2y^{2} - 125 = 0.$$

Section 7.5 Exercises 7.5 节习题

For the following exercises, determine the equation of the parabola using the information given.

255.

Focus $\left( {4,0} \right)$ and directrix $x = -4$

256\.

Focus $\left( {0,-3} \right)$ and directrix $y = 3$

257.

Focus $\left( {0,0.5} \right)$ and directrix $y = -0.5$

258\.

Focus $\left( {2,\ 3} \right)$ and directrix $x = -2$

259.

Focus $\left( {0,2} \right)$ and directrix $y = 4$

260\.

Focus $\left( {-1,4} \right)$ and directrix $x = 5$

261.

Focus $\left( {-3,5} \right)$ and directrix $y = 1$

262\.

Focus $\left( {\frac{5}{2},-4} \right)$ and directrix $x = \frac{7}{2}$

For the following exercises, determine the equation of the ellipse using the information given.

263.

Endpoints of major axis at $\left( {4,0} \right),\left( {-4,0} \right)$ and foci located at $\left( {2,0} \right),\left( {-2,0} \right)$

264\.

Endpoints of major axis at $\left( {0,5} \right),\left( {0,-5} \right)$ and foci located at $\left( {0,3} \right),\left( {0,-3} \right)$

265.

Endpoints of minor axis at $\left( {0,2} \right),\left( {0,-2} \right)$ and foci located at $\left( {3,0} \right),\left( {-3,0} \right)$

266\.

Endpoints of major axis at $\left( {-3,3} \right),\left( {7,3} \right)$ and foci located at $\left( {-2,3} \right),\left( {6,3} \right)$

267.

Endpoints of major axis at $\left( {-3,5} \right),\left( {-3,-3} \right)$ and foci located at $\left( {-3,3} \right),\left( {-3,-1} \right)$

268\.

Endpoints of minor axis at $\left( {0,0} \right),\left( {0,4} \right)$ and foci located at $\left( {5,2} \right),\left( {-5,2} \right)$

269.

Foci located at $\left( {2,0} \right),\ \left( {-2,0} \right)$ and eccentricity of $\frac{1}{2}$

270\.

Foci located at $\left( {0,-3} \right),\ \left( {0,3} \right)$ and eccentricity of $\frac{3}{4}$

For the following exercises, determine the equation of the hyperbola using the information given.

271.

Vertices located at $\left( {5,0} \right),\left( {-5,0} \right)$ and foci located at $\left( {6,0} \right),\left( {-6,0} \right)$

272\.

Vertices located at $\left( {0,2} \right),\left( {0,-2} \right)$ and foci located at $\left( {0,3} \right),\left( {0,-3} \right)$

273.

Endpoints of the conjugate axis located at $\left( {0,3} \right),\left( {0,-3} \right)$ and foci located $\left( {4,0} \right),\left( {-4,0} \right)$

274\.

Vertices located at $\left( {0,1} \right),\left( {6,1} \right)$ and focus located at $\left( {8,1} \right)$

275.

Vertices located at $\left( {-2,0} \right),\left( {-2,-4} \right)$ and focus located at $\left( {-2,-8} \right)$

276\.

Endpoints of the conjugate axis located at $\left( {3,2} \right),\left( {3,4} \right)$ and focus located at $\left( {3,7} \right)$

277.

Foci located at $( - 6,0),(6,0)$ and eccentricity of 3

278\.

$\left( {0,10} \right),\left( {0,-10} \right)$ and eccentricity of 2.5

For the following exercises, consider the following polar equations of conics. Determine the eccentricity and identify the conic.

279.

$r = \frac{-1}{1 + \text{cos}\ \theta}$

280\.

$r = \frac{8}{2 - \text{sin}\ \theta}$

281.

$r = \frac{5}{2 + \text{sin}\ \theta}$

282\.

$r = \frac{5}{-1 + 2\ \text{sin}\ \theta}$

283.

$r = \frac{3}{2 - 6\ \text{sin}\ \theta}$

284\.

$r = \frac{3}{-4 + 3\ \text{sin}\ \theta}$

For the following exercises, find a polar equation of the conic with focus at the origin and eccentricity and directrix as given.

285.

$\text{Directrix:}\ x = 4;e = \frac{1}{5}$

286\.

$\text{Directrix:}\ x = -4;e = 5$

287.

$\text{Directrix: y} = 2;e = 2$

288\.

$\text{Directrix: y} = -2;e = \frac{1}{2}$

For the following exercises, sketch the graph of each conic.

289.

$r = \frac{1}{1 + \text{sin}\ \theta}$

290\.

$r = \frac{1}{1 + \text{cos}\ \theta}$

291.

$r = \frac{4}{1 + \text{cos}\ \theta}$

292\.

$r = \frac{10}{5 + 4\ \text{sin}\ \theta}$

293.

$r = \frac{15}{3 - 2\ \text{cos}\ \theta}$

294\.

$r = \frac{32}{3 + 5\ \text{sin}\ \theta}$

295.

$r(2 + \text{sin}\ \theta) = 4$

296\.

$r = \frac{3}{2 + 6\ \text{sin}\ \theta}$

297.

$r = \frac{3}{-4 + 2\ \text{sin}\ \theta}$

298\.

$\begin{array}{l} {\frac{x^{2}}{9} + \frac{y^{2}}{4} = 1} \\ \end{array}$

299.

$\frac{x^{2}}{4} + \frac{y^{2}}{16} = 1$

300\.

$4x^{2} + 9y^{2} = 36$

301.

$25x^{2} - 4y^{2} = 100$

302\.

$\frac{x^{2}}{16} - \frac{y^{2}}{9} = 1$

303.

$x^{2} = 12y$

304\.

$y^{2} = 20x$

305.

$12x = 5y^{2}$

For the following equations, determine which of the conic sections is described.

306\.

$xy = 4$

307.

$x^{2} + 4xy - 2y^{2} - 6 = 0$

308\.

$x^{2} + 2\sqrt{3}xy + 3y^{2} - 6 = 0$

309.

$x^{2} - xy + y^{2} - 2 = 0$

310\.

$34x^{2} - 24xy + 41y^{2} - 25 = 0$

311.

$52x^{2} - 72xy + 73y^{2} + 40x + 30y - 75 = 0$

312\.

The mirror in an automobile headlight has a parabolic cross section, with the lightbulb at the focus. On a schematic, the equation of the parabola is given as $x^{2} = 4y.$ At what coordinates should you place the lightbulb?

313.

A satellite dish is shaped like a paraboloid of revolution. The receiver is to be located at the focus. If the dish is 12 feet across at its opening and 4 feet deep at its center, where should the receiver be placed?

314\.

Consider the satellite dish of the preceding problem. If the dish is 8 feet across at the opening and 2 feet deep, where should we place the receiver?

315.

A searchlight is shaped like a paraboloid of revolution. A light source is located 1 foot from the base along the axis of symmetry. If the opening of the searchlight is 3 feet across, find the depth.

316\.

Whispering galleries are rooms designed with elliptical ceilings. A person standing at one focus can whisper and be heard by a person standing at the other focus because all the sound waves that reach the ceiling are reflected to the other person. If a whispering gallery has a length of 120 feet and the foci are located 30 feet from the center, find the height of the ceiling at the center.

317.

A person is standing 8 feet from the nearest wall in a whispering gallery. If that person is at one focus and the other focus is 80 feet away, what is the length and the height at the center of the gallery?

For the following exercises, determine the polar equation form of the orbit given the length of the major axis and eccentricity for the orbits of the comets or planets. Distance is given in astronomical units (AU).

318\.

Halley’s Comet: length of major axis = 35.88, eccentricity = 0.967

319.

Hale-Bopp Comet: length of major axis = 525.91, eccentricity = 0.995

320\.

Mars: length of major axis = 3.049, eccentricity = 0.0934

321.

Jupiter: length of major axis = 10.408, eccentricity = 0.0484

Key Terms 关键术语

angular coordinate

$\theta$ the angle formed by a line segment connecting the origin to a point in the polar coordinate system with the positive radial (*x*) axis, measured counterclockwise

cardioid

a plane curve traced by a point on the perimeter of a circle that is rolling around a fixed circle of the same radius; the equation of a cardioid is $r = a\left( {1 + \text{sin}\ \theta} \right)$ or $r = a\left( {1 + \text{cos}\ \theta} \right)$

conic section

a conic section is any curve formed by the intersection of a plane with a cone of two nappes

cusp

a pointed end or part where two curves meet

cycloid

the curve traced by a point on the rim of a circular wheel as the wheel rolls along a straight line without slippage

directrix

a directrix (plural: directrices) is a line used to construct and define a conic section; a parabola has one directrix; ellipses and hyperbolas have two

discriminant

the value $4AC - B^{2},$ which is used to identify a conic when the equation contains a term involving $xy,$ is called a discriminant

eccentricity

the eccentricity is defined as the distance from any point on the conic section to its focus divided by the perpendicular distance from that point to the nearest directrix

focal parameter

the focal parameter is the distance from a focus of a conic section to the nearest directrix

focus

a focus (plural: foci) is a point used to construct and define a conic section; a parabola has one focus; an ellipse and a hyperbola have two

general form

an equation of a conic section written as a general second-degree equation

limaçon

the graph of the equation $r = a + b\ \text{sin}\ \theta$ or $r = a + b\ \text{cos}\ \theta.$ If $a = b$ then the graph is a cardioid

major axis

the major axis of a conic section passes through the vertex in the case of a parabola or through the two vertices in the case of an ellipse or hyperbola; it is also an axis of symmetry of the conic; also called the transverse axis

minor axis

the minor axis is perpendicular to the major axis and intersects the major axis at the center of the conic, or at the vertex in the case of the parabola; also called the conjugate axis

nappe

a nappe is one half of a double cone

orientation

the direction that a point moves on a graph as the parameter increases

parameter

an independent variable that both *x* and *y* depend on in a parametric curve; usually represented by the variable *t*

parameterization of a curve

rewriting the equation of a curve defined by a function $y = f(x)$ as parametric equations

parametric curve

the graph of the parametric equations $x(t)$ and $y(t)$ over an interval $a \leq t \leq b$ combined with the equations

parametric equations

the equations $x = x(t)$ and $y = y(t)$ that define a parametric curve

polar axis

the horizontal axis in the polar coordinate system corresponding to $r \geq 0$

polar coordinate system

a system for locating points in the plane. The coordinates are $r,$ the radial coordinate, and $\theta,$ the angular coordinate

polar equation

an equation or function relating the radial coordinate to the angular coordinate in the polar coordinate system

pole

the central point of the polar coordinate system, equivalent to the origin of a Cartesian system

radial coordinate

$r$ the coordinate in the polar coordinate system that measures the distance from a point in the plane to the pole

rose

graph of the polar equation $r = a\ \text{cos}\ 2\theta$ or $r = a\ \text{sin}\ 2\theta$ for a positive constant *a*

space-filling curve

a curve that completely occupies a two-dimensional subset of the real plane

standard form

an equation of a conic section showing its properties, such as location of the vertex or lengths of major and minor axes

vertex

a vertex is an extreme point on a conic section; a parabola has one vertex at its turning point. An ellipse has two vertices, one at each end of the major axis; a hyperbola has two vertices, one at the turning point of each branch

Key Equations 关键公式

Key Concepts 关键概念

7.1 Parametric Equations 7.1 参数方程

• Parametric equations provide a convenient way to describe a curve. A parameter can represent time or some other meaningful quantity.
• It is often possible to eliminate the parameter in a parameterized curve to obtain a function or relation describing that curve.
• There is always more than one way to parameterize a curve.
• Parametric equations can describe complicated curves that are difficult or perhaps impossible to describe using rectangular coordinates.

• 参数方程提供了一种描述曲线的便捷方式。参数可以表示时间或其他有意义的量。
• 对参数化曲线常常可以消去参数，从而得到描述该曲线的函数或关系式。
• 参数化一条曲线总不止一种方式。
• 参数方程能够描述复杂的曲线，这些曲线用直角坐标难以描述，或许根本无法描述。

7.2 Calculus of Parametric Curves 7.2 参数曲线的微积分

• The derivative of the parametrically defined curve $x = x(t)$ and $y = y(t)$ can be calculated using the formula $\frac{dy}{dx} = \frac{y^{\prime}(t)}{x^{\prime}(t)}.$ Using the derivative, we can find the equation of a tangent line to a parametric curve.
• The area between a parametric curve and the *x*-axis can be determined by using the formula $A = {\int_{t_{1}}^{t_{2}}{y(t)x^{\prime}(t)\ dt}}.$
• The arc length of a parametric curve can be calculated by using the formula $s = {\int_{t_{1}}^{t_{2}}{\sqrt{\left( \frac{dx}{dt} \right)^{2} + \left( \frac{dy}{dt} \right)^{2}}dt}}.$
• The surface area of a volume of revolution revolved around the *x*-axis is given by $S = 2\pi{\int_{a}^{b}{y(t)\sqrt{\left( {x^{\prime}(t)} \right)^{2} + \left( {y^{\prime}(t)} \right)^{2}}dt}}.$ If the curve is revolved around the *y*-axis, then the formula is $S = 2\pi{\int_{a}^{b}{x(t)\sqrt{\left( {x^{\prime}(t)} \right)^{2} + \left( {y^{\prime}(t)} \right)^{2}}dt}}.$

• 由参数方程定义的曲线 $x = x(t)$ 与 $y = y(t)$ 的导数可用公式 $\frac{dy}{dx} = \frac{y^{\prime}(t)}{x^{\prime}(t)}.$ 计算。利用导数，我们可以求出参数曲线的切线方程。
• 参数曲线与 *x* 轴之间的面积可由公式 $A = {\int_{t_{1}}^{t_{2}}{y(t)x^{\prime}(t)\ dt}}.$ 确定。
• 参数曲线的弧长可由公式 $s = {\int_{t_{1}}^{t_{2}}{\sqrt{\left( \frac{dx}{dt} \right)^{2} + \left( \frac{dy}{dt} \right)^{2}}dt}}.$ 计算。
• 绕 *x* 轴旋转所得旋转体曲面的面积由 $S = 2\pi{\int_{a}^{b}{y(t)\sqrt{\left( {x^{\prime}(t)} \right)^{2} + \left( {y^{\prime}(t)} \right)^{2}}dt}}.$ 给出。若曲线绕 *y* 轴旋转，则公式为 $S = 2\pi{\int_{a}^{b}{x(t)\sqrt{\left( {x^{\prime}(t)} \right)^{2} + \left( {y^{\prime}(t)} \right)^{2}}dt}}.$

7.3 Polar Coordinates 7.3 极坐标

• The polar coordinate system provides an alternative way to locate points in the plane.
• Convert points between rectangular and polar coordinates using the formulas

• 极坐标系提供了在平面上确定点的另一种方式。
• 利用以下公式在直角坐标与极坐标之间转换点：

$$x = r\ \text{cos}\ \theta\ \text{and}\ y = r\ \text{sin}\ \theta$$

and

$$r = \sqrt{x^{2} + y^{2}}\ \text{and}\ \text{tan}\ \theta = \frac{y}{x}.$$

• To sketch a polar curve from a given polar function, make a table of values and take advantage of periodic properties.
• Use the conversion formulas to convert equations between rectangular and polar coordinates.
• Identify symmetry in polar curves, which can occur through the pole, the horizontal axis, or the vertical axis.

• 要根据给定的极坐标函数描绘极坐标曲线，可建立数值表并利用周期性。
• 利用转换公式在直角坐标与极坐标之间转换方程。
• 识别极坐标曲线中的对称性，它可能关于极点、水平轴或垂直轴出现。

7.4 Area and Arc Length in Polar Coordinates 7.4 极坐标中的面积与弧长

• The area of a region in polar coordinates defined by the equation $r = f(\theta)$ with $\alpha \leq \theta \leq \beta$ is given by the integral $A = \frac{1}{2}{\int_{\alpha}^{\beta}\left\lbrack {f(\theta)} \right\rbrack}^{2}d\theta.$
• To find the area between two curves in the polar coordinate system, first find the points of intersection, then subtract the corresponding areas.
• The arc length of a polar curve defined by the equation $r = f(\theta)$ with $\alpha \leq \theta \leq \beta$ is given by the integral $L = {\int_{\alpha}^{\beta}{\sqrt{\left\lbrack {f(\theta)} \right\rbrack^{2} + \left\lbrack {f^{\prime}(\theta)} \right\rbrack^{2}}d\theta = {\int_{\alpha}^{\beta}{\sqrt{r^{2} + \left( \frac{dr}{d\theta} \right)^{2}}d\theta}}}}.$

• 由方程 $r = f(\theta)$（$\alpha \leq \theta \leq \beta$）定义的极坐标区域的面积由积分 $A = \frac{1}{2}{\int_{\alpha}^{\beta}\left\lbrack {f(\theta)} \right\rbrack}^{2}d\theta.$ 给出。
• 要求极坐标中两曲线之间的面积，先求出交点，再相减相应的面积。
• 由方程 $r = f(\theta)$（$\alpha \leq \theta \leq \beta$）定义的极坐标曲线的弧长由积分 $L = {\int_{\alpha}^{\beta}{\sqrt{\left\lbrack {f(\theta)} \right\rbrack^{2} + \left\lbrack {f^{\prime}(\theta)} \right\rbrack^{2}}d\theta = {\int_{\alpha}^{\beta}{\sqrt{r^{2} + \left( \frac{dr}{d\theta} \right)^{2}}d\theta}}}}.$ 给出。

7.5 Conic Sections 7.5 圆锥曲线

• The equation of a vertical parabola in standard form with given focus and directrix is $y = \frac{1}{4p}\left( {x - h} \right)^{2} + k$ where *p* is the distance from the vertex to the focus and $\left( {h,k} \right)$ are the coordinates of the vertex.
• The equation of a horizontal ellipse in standard form is $\frac{\left( {x - h} \right)^{2}}{a^{2}} + \frac{\left( {y - k} \right)^{2}}{b^{2}} = 1$ where the center has coordinates $\left( {h,k} \right),$ the major axis has length 2*a,* the minor axis has length 2*b*, and the coordinates of the foci are $\left( {h \pm c,k} \right),$ where $c^{2} = a^{2} - b^{2}.$
• The equation of a horizontal hyperbola in standard form is $\frac{\left( {x - h} \right)^{2}}{a^{2}} - \frac{\left( {y - k} \right)^{2}}{b^{2}} = 1$ where the center has coordinates $\left( {h,k} \right),$ the vertices are located at $\left( {h \pm a,k} \right),$ and the coordinates of the foci are $\left( {h \pm c,k} \right),$ where $c^{2} = a^{2} + b^{2}.$
• The eccentricity of an ellipse is less than 1, the eccentricity of a parabola is equal to 1, and the eccentricity of a hyperbola is greater than 1. The eccentricity of a circle is 0.
• The polar equation of a conic section with eccentricity *e* is $r = \frac{ep}{1 \pm e\ \text{cos}\ \theta}$ or $r = \frac{ep}{1 \pm e\ \text{sin}\ \theta},$ where *p* represents the focal parameter.
• To identify a conic generated by the equation $Ax^{2} + Bxy + Cy^{2} + Dx + Ey + F = 0,$ first calculate the discriminant $D = 4AC - B^{2}.$ If $D > 0$ then the conic is an ellipse, if $D = 0$ then the conic is a parabola, and if $D < 0$ then the conic is a hyperbola.

• 给定焦点与准线、开口向上的抛物线的标准形方程为 $y = \frac{1}{4p}\left( {x - h} \right)^{2} + k$，其中 *p* 为顶点到焦点的距离，$\left( {h,k} \right)$ 为顶点坐标。
• 水平椭圆的标准形方程为 $\frac{\left( {x - h} \right)^{2}}{a^{2}} + \frac{\left( {y - k} \right)^{2}}{b^{2}} = 1$，其中中心坐标为 $\left( {h,k} \right)$，长轴长为 2*a*，短轴长为 2*b*，焦点坐标为 $\left( {h \pm c,k} \right)$，且 $c^{2} = a^{2} - b^{2}.$
• 水平双曲线的标准形方程为 $\frac{\left( {x - h} \right)^{2}}{a^{2}} - \frac{\left( {y - k} \right)^{2}}{b^{2}} = 1$，其中中心坐标为 $\left( {h,k} \right)$，顶点位于 $\left( {h \pm a,k} \right)$，焦点坐标为 $\left( {h \pm c,k} \right)$，且 $c^{2} = a^{2} + b^{2}.$
• 椭圆的离心率小于 1，抛物线的离心率等于 1，双曲线的离心率大于 1。圆的离心率为 0。
• 离心率为 *e* 的圆锥曲线的极坐标方程为 $r = \frac{ep}{1 \pm e\ \text{cos}\ \theta}$ 或 $r = \frac{ep}{1 \pm e\ \text{sin}\ \theta}$，其中 *p* 表示焦参数。
• 要判别由方程 $Ax^{2} + Bxy + Cy^{2} + Dx + Ey + F = 0$ 生成的圆锥曲线，先计算判别式 $D = 4AC - B^{2}.$ 若 $D > 0$，则为椭圆；若 $D = 0$，则为抛物线；若 $D < 0$，则为双曲线。

Review Exercises 复习题

*True or False?* Justify your answer with a proof or a counterexample.

322\.

The rectangular coordinates of the point $\left( {4,\frac{5\pi}{6}} \right)$ are $\left( {2\sqrt{3},-2} \right).$

323.

The equations $x = \text{cosh}(3t),$ $y = 2\ \text{sinh}(3t)$ represent a hyperbola.

324\.

The arc length of the spiral given by $r = \frac{\theta}{2}$ for $0 \leq \theta \leq 3\pi$ is $\frac{9}{4}\pi^{3}.$

325.

Given $x = f(t)$ and $y = g(t),$ if $\frac{dx}{dy} = \frac{dy}{dx},$ then $f(t) = g(t) + \text{C,}$ where C is a constant.

For the following exercises, sketch the parametric curve and eliminate the parameter to find the Cartesian equation of the curve.

326\.

$x = 1 + t,$ $y = t^{2} - 1,$ $-1 \leq t \leq 1$

327.

$x = e^{t},$ $y = 1 - e^{3t},$ $0 \leq t \leq 1$

328\.

$x = \text{sin}\ \theta,$ $y = 1 - \text{csc}\ \theta,$ $0 \leq \theta \leq 2\pi$

329.

$x = 4\ \text{cos}\ \phi,$ $y = 1 - \text{sin}\ \phi,$ $0 \leq \phi \leq 2\pi$

For the following exercises, sketch the polar curve and determine what type of symmetry exists, if any.

330\.

$r = 4\ \text{sin}\left( \frac{\theta}{3} \right)$

331.

$r = 5\ \text{cos}\left( {5\theta} \right)$

For the following exercises, find the polar equation for the curve given as a Cartesian equation.

332\.

$x + y = 5$

333.

$y^{2} = 4 + x^{2}$

For the following exercises, find an equation of the tangent line to the given curve. Graph both the function and its tangent line.

334\.

$x = \text{ln}(t),$ $y = t^{2} - 1,$ $t = 1$

335.

$r = 3 + \text{cos}\left( {2\theta} \right),$ $\theta = \frac{3\pi}{4}$

336\.

Find $\frac{dy}{dx},$ $\frac{dx}{dy},$ and $\frac{d^{2}x}{dy^{2}}$ of $y = \left( {2 + e^{\text{−}t}} \right),$ $x = 1 - \text{sin}(t)$

For the following exercises, find the area of the region.

337.

$x = t^{2},$ $y = \text{ln}(t),$ $0 \leq t \leq e$

338\.

$r = 1 - \text{sin}\ \theta$ in the first quadrant

For the following exercises, find the arc length of the curve over the given interval.

339.

$x = 3t + 4,$ $y = 9t - 2,$ $0 \leq t \leq 3$

340\.

$r = 6\ \text{cos}\ \theta,$ $0 \leq \theta \leq 2\pi.$ Check your answer by geometry.

For the following exercises, find the Cartesian equation describing the given shapes.

341.

A parabola with focus $(2,-5)$ and directrix $x = 6$

342\.

An ellipse with a major axis length of 10 and foci at $\left( {-7,2} \right)$ and $\left( {1,2} \right)$

343.

A hyperbola with vertices at $(–2,3)$ and $(-2,-5)$ and foci at $(-2,-6)$ and $(-2,4)$

For the following exercises, determine the eccentricity and identify the conic. Sketch the conic.

344\.

$r = \frac{6}{1 + 3\ \text{cos}(\theta)}$

345.

$r = \frac{4}{3 - 2\ \text{cos}\ \theta}$

346\.

$r = \frac{7}{5 - 5\ \text{cos}\ \theta}$

347.

Determine the Cartesian equation describing the orbit of Pluto, the most eccentric orbit around the Sun. The length of the major axis is 39.26 AU and minor axis is 38.07 AU. What is the eccentricity?

348\.

The C/1980 E1 comet was observed in 1980. Given an eccentricity of 1.057 and a perihelion (point of closest approach to the Sun) of 3.364 AU, find the Cartesian equations describing the comet’s trajectory. Are we guaranteed to see this comet again? (*Hint*: Consider the Sun at point $(0,0).)$