Chapter 8.5, Problem 3E

Solution

To find:The polar equation for the conic with a focus at the pole, eccentricity and directrix. Also, to identify the conic and graph it.

The equation of the conic is $r=\frac{12}{5+3\sin \theta }$ and the conic is anellipse.

Given information:

The given eccentricity is $e=\frac{3}{5}$ , and directrix is $y=4$ .

Formula used:

  $e<1$ , the conic is an ellipse.

  $e=1$ , the conic is a parabola.

  $e>1$ , the conic is a hyperbola.

  $r=\frac{ke}{1+e\cos \theta }$ when directrix is vertical, to the right of the pole: $x=k$ .

  $r=\frac{ke}{1-e\cos \theta }$ when directrix is vertical, to the right of the pole: $x=-k$ .

  $r=\frac{ke}{1+e\sin \theta }$ when directrix is horizontal, to the right of the pole: $y=k$ .

  $r=\frac{ke}{1-e\sin \theta }$ when directrix is horizontal, to the right of the pole: $y=-k$ .

Calculation:

The value of eccentricity is $e=\frac{3}{5}$ .Since $e<1$ , the conic is an ellipse.

The direction equation is $y=4$ , therefore the equation to be used is $r=\frac{ke}{1+e\sin \theta }$ .

Compare $y=k$ with $y=4$ .

Therefore, $k=4$ .

Substitute $\frac{3}{5}$ for $e$ , and $4$ for $k$ in $r=\frac{ke}{1+e\sin \theta }$ .

  $\begin{array}{l}r=\frac{ke}{1+e\sin \theta }\\ r=\frac{4\times \left(\frac{3}{4}\right)}{1-\frac{3}{4}\sin \theta }\\ r=\frac{12}{5+3\sin \theta }\end{array}$

Plot the graph for the equation $r=\frac{12}{5+3\sin \theta }$ .

  

Therefore, the equation of the conic is $r=\frac{12}{5+3\sin \theta }$ and the conic is an ellipse.