Chapter 9, Problem 101CR

To determine

To find:

The polar equation of the conic with its focus at the pole.

Answer to Problem 101CR

The polar equation of the conic is $r=\frac{15}{3+2\cos \theta }$ .

Explanation of Solution

Given information:

The conic is ellipse and its vertex at $\left(5,\frac{\pi }{2}\right),\left(1,\frac{3\pi }{2}\right)$

Calculation:

For the given parabola the vertex is given as,

$\left(r,\theta \right)=\left(5,\frac{\pi }{2}\right)\text{ or }\left(1,\frac{3\pi }{2}\right)$

Because the directrix is vertical and to the right of the pole the equation of the conic section has the form,

$r=\frac{ep}{1+e\cos \theta }$

The length of the major axis is the distance between the 2 vertices, $2a$

So,

$\begin{array}{l}2a=1+5\\ a=\frac{6}{2}\\ a=3\end{array}$

The distance from center to focus $\left(0,0\right)$ is $c$

Here, $c=2$

The eccentricity $\left(e\right)=\frac{c}{a}=\frac{2}{3}$

Substitute the equation,

$\begin{array}{l}r=\frac{ep}{1+e\cos \theta }\\ r=\frac{\left(\frac{2}{3}\right)p}{1+\left(\frac{2}{3}\right)\cos \theta }\\ r=\frac{2p}{3+2\cos \theta }\left[\text{multiply both numerator and denominator by 3}\right]\end{array}$

To get the value of $\text{'}p\text{'}$ put any of the given vertex into the equation,

$\begin{array}{l}r=\frac{2p}{3+2\cos \theta }\\ 5=\frac{2p}{3+2\cos \left(\frac{\pi }{2}\right)}\\ 5=\frac{2p}{3+2\times 0}\\ 2p=15\\ p=\frac{15}{2}\end{array}$

Hence, the required equation of the conic section is,

$\begin{array}{l}r=\frac{\left(\frac{2}{3}\right)\cdot \left(\frac{15}{2}\right)}{1+\left(\frac{2}{3}\right)\cos \theta }\\ r=\frac{5}{1+\left(\frac{2}{3}\right)\cos \theta }\\ r=\frac{15}{3+2\cos \theta }\left[\text{multiply both numerator and denominator by 3}\right]\end{array}$

Therefore, the polar equation of the conic is $r=\frac{15}{3+2\cos \theta }$ .