Chapter 8.5, Problem 20E

Solution

To find:The graph for the given polar equation and the viewing window of the graph.

Theconic is a parabola, and graph (a) with the viewing window of $\left[-3,3\right]$ by $\left[-6,6\right]$ is represented by the polar equation $r=\frac{15}{4+4\sin \theta }$ .

Given information:

The given polar equation is $r=\frac{15}{4+4\sin \theta }$ .

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:

Compare $r=\frac{15}{4+4\sin \theta }$ with $r=\frac{ke}{1-e\sin \theta }$ .

Take $4$ common in the denominator.

  $r=\frac{\frac{15}{4}}{1+\cos \theta }$

Here, $k=\frac{15}{4}$ and $e=1$ .

Since, $e=1$ , the conic is aparabola.

The equation of directrix is $y=k$ .

Substitute $\frac{15}{4}$ for $k$ in $y=k$ .

Thus, equation of directrix for $r=\frac{15}{4+4\sin \theta }$ is $y=\frac{15}{4}$ .

This matches the graph (a).

  

The viewing window of the graph is $\left[-3,3\right]$ by $\left[-6,6\right]$ .

Therefore, the conic is aparabola, and graph (a) with the viewing window of $\left[-3,3\right]$ by $\left[-6,6\right]$ is represented by the polar equation $r=\frac{15}{4+4\sin \theta }$ .