Chapter 11.6, Problem 26E

a.

To determine

To show: that the conic is hyperbola.

Explanation of Solution

Given information:

The polar equation is $r=\frac{10}{1-4\sin \theta }$ .

Proof: since, the polar equation in standard form is $r=\frac{\left(4\right)\left(\frac{5}{2}\right)}{1-4\sin \theta }$ ,

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,

$\theta$	$r$
$\frac{\pi }{2}$	$-\frac{10}{3}$
$0$	$10$
$\frac{5\pi }{2}$	$-\frac{10}{3}$

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 $r=\frac{10}{1-4\sin \theta }$ .

Proof: since, the polar equation in standard form is $r=\frac{\left(4\right)\left(\frac{5}{2}\right)}{1-4\sin \theta }$ ,

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,

$\theta$	$r$
$\frac{\pi }{2}$	$-\frac{10}{3}$
$0$	$10$
$\frac{5\pi }{2}$	$-\frac{10}{3}$

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 $\left(0,-2\right),\left(0,-3.33\right)$ and directrix is $y=-\frac{5}{2}$ .

c.

To determine

To find : the centre of the hyperbola.

Answer to Problem 26E

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

Explanation of Solution

Given information :

The polar equation is $r=\frac{10}{1-4\sin \theta }$ .

Calculation : from the graph in the part (b) it can be observed that the centre of hyperbola is $\left(0,-2.66\right)$ .