Chapter 10, Problem 54RE

To determine

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

Answer to Problem 54RE

The conic is parabola focus at pole.

Explanation of Solution

Given:

  $\text{r=}\frac{6}{\text{1+sinθ}}$ .

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:

  $\text{r=}\frac{6}{\text{1+sinθ}}$

Comparing this with polar equation:

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

Here $\text{e=1 and p=6}$ .

So, the conic is parabola focus at pole.

The directrix is parallel to polar axis at a distance 6 units above of the pole

Vertex of the parabola is

  $\left(\text{3,π}\right)$ .

The graph of the polar equation $\text{r=}\frac{6}{\text{1+sinθ}}$ .

  

Hence, the conic is parabola focus at pole.