Chapter 11.6, Problem 27E

a.

To determine

To show: that the conic is hyperbola.

Explanation of Solution

Given information:

The polar equation is $r=\frac{20}{2-3\sin \theta }$ .

Proof: since, the polar equation in standard form is $r=\frac{\left(\frac{3}{2}\right)\left(\frac{20}{3}\right)}{1-\frac{3}{2}\sin \theta }$ ,

Form the polar equation it can be observed that, $e=4\text{ and }d=\frac{5}{2}$ .

Since, $e>1$ ,

Thus, the conic is hyperbola.

Now, by using plotting point method,

$\theta$	$r$
$\frac{\pi }{2}$	$-20$
$0$	$10$
$\frac{5\pi }{2}$	$-20$

Now, using the above table and join the points,

The graph can be obtained as:

  

b.

To determine

To draw: the hyperbola with its directrix and vertex.

Explanation of Solution

Given information:

The polar equation is $r=\frac{20}{2-3\sin \theta }$ .

Proof: since, the polar equation in standard form is $r=\frac{\left(\frac{3}{2}\right)\left(\frac{20}{3}\right)}{1-\frac{3}{2}\sin \theta }$ ,

Form the polar equation it can be observed that, $e=4\text{ and }d=\frac{5}{2}$ .

Since, $e>1$ ,

Thus, the conic is hyperbola.

Now, by using plotting point method,

$\theta$	$r$
$\frac{\pi }{2}$	$-20$
$0$	$10$
$\frac{5\pi }{2}$	$-20$

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 parabola is $\left(4,\frac{3\pi }{2}\right),\left(-20,\frac{\pi }{2}\right)$ and directrix is $y=-\frac{20}{3}$ .

c.

To determine

To find : the centre of the hyperbola.

Answer to Problem 27E

the centre of the hyperbola is $\left(0,-2.66\right)$ .

Explanation of Solution

Given information :

The polar equation is $r=\frac{20}{2-3\sin \theta }$ .

Calculation : from the graph in the part (b) it can be observed that the centre of hyperbola is $\left(12,\frac{3\pi }{2}\right)$ .