Chapter 11.6, Problem 35E

a.

To determine

To find: the eccentricity and identify the conic.

Answer to Problem 35E

The eccentricity is $\frac{5}{2}$ and the conic is hyperbola.

Explanation of Solution

Given information:

The polar equation is $r=\frac{7}{2-5\sin \theta }$ .

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,

  $r=\frac{\frac{5}{2}\left(\frac{7}{5}\right)}{1-\frac{5}{2}\sin \theta }$

By comparing both polar equation,

  $e=\frac{5}{2}\text{ and }d=\frac{7}{5}$

Thus, the eccentricity of the conic is $\frac{5}{2}$ .

Because the $e>1$ ,

Thus, the conic is hyperbola.

b.

To determine

To draw: the conic.

Explanation of Solution

Given information:

The polar equation is $r=\frac{7}{2-5\sin \theta }$ .

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

$\theta$	$r$
$\frac{\pi }{2}$	$-\frac{7}{3}$
$0$	$\frac{7}{2}$
$\frac{5\pi }{2}$	$-\frac{7}{3}$

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 conic is $\left(0,-1\right),\left(0,-2.33\right)$ .