Chapter 11.6, Problem 38E

a.

To determine

To find: the eccentricity and directrix of the conic.

Answer to Problem 38E

The eccentricity is $\frac{3}{5}$ and the directrix of the conic is $x=-\frac{2}{3}$ .

Explanation of Solution

Given information:

The polar equation is $r=\frac{2}{5-3\cos \theta },\theta =\frac{2\pi }{3}$ .

Calculation: to find the eccentricity and directrix, first compare the polar equation with the standard equation,

The standard equation is $r=\frac{ed}{1-e\cos \theta }$ .

Now, convert the polar equation in form of standard form of equation,

  $r=\frac{\frac{3}{5}\left(\frac{2}{3}\right)}{1-\frac{3}{5}\cos \theta }$

By comparing both polar equation,

  $e=\frac{3}{5}\text{ and }d=\frac{2}{3}$

Thus, the eccentricity of the conic is $\frac{3}{4}$ .

And the directrix of the conic is $x=-\frac{2}{3}$ .

b.

To determine

To find: the polar equation after rotating the conic.

Answer to Problem 38E

The polar equation of conic is $r=\frac{2}{5-3\cos \left(\theta -\frac{2\pi }{3}\right)}$ after rotating.

Explanation of Solution

Given information:

The polar equation is $r=\frac{2}{5-3\cos \theta },\theta =\frac{2\pi }{3}$ .

Calculation: to find the polar equation, substitute $\theta =\theta -\frac{2\pi }{3}$ in the equation,

  $r=\frac{2}{5-3\cos \left(\theta -\frac{2\pi }{3}\right)}\text{ [substitute]}$

Thus, the polar equation is $r=\frac{2}{5-3\cos \left(\theta -\frac{2\pi }{3}\right)}$ .

c.

To determine

To draw: the original conic and after rotating.

Explanation of Solution

Given information:

The polar equation is $r=\frac{2}{5-3\cos \theta },\theta =\frac{2\pi }{3}$ .

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

$\theta$	$r$
$0$	$1$
$\frac{\pi }{2}$	$\frac{2}{5}$
$\pi$	$\frac{1}{4}$

By using the above table, and the graph of the conic after rotating can be obtained as:

  

Interpretation: from the above graph it can be observed that the rotating graph can be obtained by rotate the original graph $\frac{2\pi }{3}$ .