Chapter 11, Problem 65RE

a.

To determine

To find: The eccentricity and identify the given conic.

Answer to Problem 65RE

The eccentricity of the given polar equation of a conic is $e=2$

Explanation of Solution

Given information:

The polar equation of the conic is

  $\text{r =}\frac{\text{A}}{\text{1+2sinθ}}$

Concept used:

The general equation of a conic in the polar form is

  $\text{r =}\frac{k}{\text{1}\pm \text{esinθ}}$ or $\text{r =}\frac{k}{\text{1}\pm \text{e coθ}}$

Where

  $e$ is the eccentricity of the conic

If

  $e=1$ then the conic represents parabola

  $0<e<1$ then conic represents ellipse

  $e>1$ then the conic represents the hyperbola

Calculation:

Given polar equation is $\text{r =}\frac{\text{A}}{\text{1+2sinθ}}$

Converting the given polar equation to the general form as above by the suitable rearrangement.

The required polar equation represents general form is

  $\text{r =}\frac{\text{A}}{\text{1+2sinθ}}$

Comparing with the general form find $\text{e}$ which is eccentricity

  $e=2$

Conclusion:

The eccentricity is obtained as $e>0$

Therefore, the given polar equation represents a hyperbola.

b.

To determine

To sketch: The conic and label the vertex.

Answer to Problem 65RE

From the graph it can be observed that the vertices of the conic is/are $\left(0,1.33\right)\&\left(0,4\right)$ .

Explanation of Solution

Given information:

Polar equation of the conic is

  $\text{r =}\frac{\text{A}}{\text{1+2sinθ}}$

Graph:

Sketch the graph using graphing utility.

Step 1: Press WINDOW button to access the Window editor.

Step 2: Press $\boxed{\text{Y=}}$ button.

Step 3: Enter the expression $\text{r =}\frac{\text{A}}{\text{1+2sinθ}}$ which is required to graph.

Step 4: Press GRAPH button to graph the function and adjust the windows according to the graph.

The graph is obtained as:

  

Interpretation:

From the above graph it can be observed that the vertices of the conic are $\left(0,1.33\right)\&\left(0,4\right)$