Chapter 11.6, Problem 33E

a.

To determine

To find: the eccentricity and identify the conic.

Answer to Problem 33E

The eccentricity is $\frac{1}{2}$ and the conic is ellipse.

Explanation of Solution

Given information:

The polar equation is $r=\frac{6}{2+\sin \theta }$ .

Calculation: to find the eccentricity and identify the conic, first compare the polar equation with the standard equation,

The standard equation is $r=\frac{ed}{1+e\sin \theta }$ .

Now, convert the polar equation in form of standard form of equation,

  $r=\frac{\frac{1}{2}\left(6\right)}{1+\frac{1}{2}\sin \theta }$

By comparing both polar equation,

  $e=\frac{1}{2}\text{ and }d=6$

Thus, the eccentricity of the conic is $\frac{1}{2}$ .

Because the $e<1$ ,

Thus, the conic is ellipse.

b.

To determine

To draw: the conic.

Explanation of Solution

Given information:

The polar equation is $r=\frac{6}{2+\sin \theta }$ .

Proof: to draw the conic, use plotting point method,

$\theta$	$r$
$\frac{\pi }{2}$	$2$
$0$	$3$
$\frac{5\pi }{2}$	$2$

Now, using the above table and join the points,

The graph can be obtained as:

  

Interpretation: from the above graph it can be observed that the vertices of the conic is $\left(-3,0\right),\left(3,0\right)$ .