Chapter 8.5, Problem 52E

Solution

To graph: $r=\frac{a\left(1-e^2\right)}{1+e\cos \theta }$ and identify the type of conic section for each value and to find what happens when $e=1$ .

The conic is a circle for $e=0$ .

The conic is an ellipse for $e=0.3$ and $e=0.7$ .

The conic is a hyperbola for $e=1.5$ and $e=3$ .

The graph is a single point as $r=0$ .

Given information:

  $a=2$

Formula used:

Calculation:

The graphs of the conic sections for $a=2$ .

Substitute $e=0$ in $r=\frac{a\left(1-e^2\right)}{1+e\cos \theta }$ and draw the graph.

  

Therefore, the conic circle is a circle.

Substitute $e=0.3$ in $r=\frac{a\left(1-e^2\right)}{1+e\cos \theta }$ and draw the graph.

  

Therefore, the conic circle is an ellipse.

Substitute $e=0.7$ in $r=\frac{a\left(1-e^2\right)}{1+e\cos \theta }$ and draw the graph.

  

Therefore, the conic circle is an ellipse.

Substitute $e=1.5$ in $r=\frac{a\left(1-e^2\right)}{1+e\cos \theta }$ and draw the graph.

  

Therefore, the conic circle is a hyperbola.

Substitute $e=3$ in $r=\frac{a\left(1-e^2\right)}{1+e\cos \theta }$ and draw the graph.

  

Therefore, the conic circle is a hyperbola.

Substitute $e=1$ in $r=\frac{a\left(1-e^2\right)}{1+e\cos \theta }$ and draw the graph.

  

The graph is a single point as $r=0$ .