Chapter 11, Problem 64RE

a.

To determine

To find: The eccentricity and identify the conic given.

Answer to Problem 64RE

The eccentricity of the given polar equation of a conic is $e=\frac{2}{3}$

Therefore the conic is a parabola

Explanation of Solution

Given information:

The polar equation of the conic is

  $\text{r =}\frac{2}{3+2\text{sinθ}}$

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 $\text{r =}\frac{2}{3+2\text{sinθ}}$

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

The required polar equation represents general form is

  $\text{r =}\frac{2}{3+2\text{sinθ}}=\frac{\left(\frac{2}{3}\right)}{\left(\frac{3}{3}\right)+\frac{2}{3}\sin \theta }=\frac{\left(\frac{2}{3}\right)}{1+\left(\frac{2}{3}\right)\sin \theta }$

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

  $e=\frac{2}{3}$

Conclusion:

The eccentricity obtained is $0<e<1$

Therefore the given polar equation represents an ellipse

b.

To determine

To sketch: The conic and label the vertex.

Answer to Problem 64RE

From the graph it can be observed that the vertices of the conic is $\left(0,0\cdot 4\right)\&\left(0,-2\right)$

Explanation of Solution

Given information:

Polar equation of the conic is

  $\text{r =}\frac{2}{3+2\text{sinθ}}$

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 $\text{r =}\frac{2}{3+2\text{sinθ}}$ 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 is/are $\left(0,0\cdot 4\right)\&\left(0,-2\right)$