Chapter 11.6, Problem 18E

a.

To determine

To show: that the conic is parabola.

Explanation of Solution

Given information:

The polar equation is $r=\frac{3}{2+2\sin \theta }$ .

Proof: since, the polar equation in standard form is $r=\frac{\frac{3}{2}}{1+\sin \theta }$ ,

Form the polar equation it can be observed that, $e=1\text{ and }d=\frac{3}{2}$ .

Since, $e=1$ ,

Thus, the conic is parabola.

Now, by using plotting point method,

$\theta$	$r$
$0$	$\frac{3}{2}$
$\frac{\pi }{2}$	$\frac{3}{4}$
$2\pi$	$\frac{3}{2}$

Now, using the above table and join the points,

The graph can be obtained as:

  

b.

To determine

To draw: the parabola with its directrix and vertex.

Explanation of Solution

Given information:

The polar equation is $r=\frac{3}{2+2\sin \theta }$ .

Proof: since, the polar equation in standard form is $r=\frac{\frac{3}{2}}{1+\sin \theta }$ ,

Form the polar equation it can be observed that, $e=1\text{ and }d=\frac{3}{2}$ .

Since, $e=1$ ,

Thus, the conic is parabola.

Now, by using plotting point method,

$\theta$	$r$
$0$	$\frac{3}{2}$
$\frac{\pi }{2}$	$\frac{3}{4}$
$2\pi$	$\frac{3}{2}$

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 vertex of the parabola is $\left(0,0.75\right)$ and directrix is $y=\frac{3}{2}$ .