Chapter 10, Problem 57RE

To determine

Tofind:the conic of the polar equation and graph the polar equation.

Answer to Problem 57RE

The conic is hyperbola with transverse axis perpendicular to directrix.

Explanation of Solution

Given:

  $\text{r=}\frac{8}{\text{4+8cosθ}}$ .

Concept used:

For a conic with eccentricity e

If $\text{0}\le \text{e}\le \text{1}$ the conic is ellipse.

If $e=1$ then the conic is parabola.

If $e>1$ ,then the conic is hyperbola.

For a conic with focus at origin and Directrix is $\text{x=±p}$ .

Then Polar equation of the conic is

  $\text{r=}\frac{\text{ep}}{\text{1±esinθ}}$ .

For a conic with focus at origin and Directrix is $\text{y=±p}$ .

Then Polar equation of the conic is

  $\text{r=}\frac{\text{ep}}{\text{1±ecosθ}}$ .

Calculation:

  $\text{r=}\frac{8}{\text{4+8cosθ}}$ .

Comparing this with polar equation:

  $\text{r=}\frac{\text{ep}}{\text{1±ecosθ}}$ .

  $\text{ep=2}$ .

  $\text{2p=2}$ .

  $p=1$ .

Here $\text{e=2 and p=1}$ .

If $e>1$ ,then the conic is hyperbola.

So, the conic is hyperbolawith transverse axis perpendicular to directrix.

The directrix is perpendicular to polar axis at a distance 1 units to the right of the pole.

The graph of the polar equation $\text{r=}\frac{8}{\text{4+8cosθ}}$ is:

  

Hence, the conic is hyperbolawith transverse axis perpendicular to directrix.