Chapter 11.6, Problem 25E

a.

To determine

To show: that the conic is hyperbola.

Explanation of Solution

Given information:

The polar equation is $r=\frac{8}{1+2\cos \theta }$ .

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

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

Since, $e>1$ ,

Thus, the conic is hyperbola.

Now, by using plotting point method,

$\theta$	$r$
$\frac{\pi }{2}$	$8$
$0$	$\frac{8}{3}$
$\frac{5\pi }{2}$	$\frac{8}{3}$

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 $r=\frac{8}{1+2\cos \theta }$ .

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

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

Since, $e>1$ ,

Thus, the conic is hyperbola.

Now, by using plotting point method,

$\theta$	$r$
$\frac{\pi }{2}$	$8$
$0$	$\frac{8}{3}$
$\frac{5\pi }{2}$	$\frac{8}{3}$

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 $\left(2.67,0\right),\left(8,\pi \right)$ and directrix is $x=4$ .

c.

To determine

To find : the centre of the hyperbola.

Answer to Problem 25E

the centre of the hyperbola is $\left(\frac{16}{3},0\right)$ .

Explanation of Solution

Given information :

The polar equation is $r=\frac{8}{1+2\cos \theta }$ .

Calculation : from the graph in the part (b) it can be observed that the centre of hyperbola is $\left(\frac{16}{3},0\right)$ .