Chapter 9.7, Problem 47E

To determine

To find: polar equation of the conic

Answer to Problem 47E

The polar equation of the conic is $\frac{10}{3+2\cos \theta }$ .

Explanation of Solution

Given:

  $\begin{array}{l}ConicVertexorVertices\\ \text{Ellipse }\left(2,0\right)\text{,}\left(10,\pi \right)\end{array}$

Calculation:

Convert the two vertices into rectangular coordinates. Find the center of the ellipse.

  $\begin{array}{l}\left(2,0\right)=\left(2,0\right)\\ \left(10,\pi \right)=\left(-10,0\right)\\ center=\left(\frac{-10+2}{2},\frac{0+0}{2}\right)=\left(-4,0\right)\end{array}$

Find a, the distance between one of the vertices and the center.

  $a=6$

Find c, the distance between the center and the foci at $\left(0,0\right)$

  $c=4$

Find the eccentricity.

  $e=\frac{c}{a}=\frac{4}{6}=\frac{2}{3}$

Let d1 be the distance between the directrix and the vertex that is closest to the focus at $\left(0,0\right)$ . Let d2 be the distance between the same vertex and the focus. $e=d2/d1$ , Find $p=d2+d1$ .

  $\begin{array}{l}d2=a-c=6-4=2\\ \frac{d2}{d1}=\frac{2}{d1}=\frac{2}{3}\\ d1=3\\ p=d2+d1=2+3=5\end{array}$

The directrix is to the right of the pole since the closest vertex to the pole is to the right of the pole. Write the equation for the ellipse.

  $\begin{array}{l}r=\frac{ep}{1+e\cos \theta }\\ r=\frac{\left(\frac{10}{3}\right)}{1+\frac{2}{3}\cos \theta }=\frac{10}{3+2\cos \theta }\end{array}$

Conclusion:

Therefore, the polar equation of the conic is $\frac{10}{3+2\cos \theta }$ .