Chapter 9.4, Problem 39PPS

a.

To determine

To graph the ellipses with given equations on same graph.

Explanation of Solution

Given:

The given equations of ellipses are

  $\frac{x^2}{81}+\frac{y^2}{36}=1$ And $\frac{x^2}{81}+\frac{y^2}{9}=1$

Calculation:

The following is the graph of both ellipses on single graph

  

b.

To determine

To describe the difference between two graphs.

Explanation of Solution

Given:

The given equations of ellipses are -

  $\frac{x^2}{81}+\frac{y^2}{36}=1$ And $\frac{x^2}{81}+\frac{y^2}{9}=1$

Calculation:

One ellipse is much flatter than the other. Both have the same major axis, but one has a much shorter minor axis.

c.

To determine

To find the eccentricity of each ellipse.

Answer to Problem 39PPS

  $\frac{\sqrt{5}}{3}$ , $\frac{2\sqrt{2}}{3}$

Explanation of Solution

Given:

The given equations of ellipses are -

  $\frac{x^2}{81}+\frac{y^2}{36}=1$ And $\frac{x^2}{81}+\frac{y^2}{9}=1$

Calculation:

The eccentricity is calculated by the following formula-

  $\begin{array}{l}\frac{c}{a}\\ c_1=\sqrt{a_1{}^2-b_1{}^2}=\sqrt{81-36}=\sqrt{45}=\pm 3\sqrt{5}\\ a_1=a_2=\sqrt{81}=\pm 9\\ c_2=\sqrt{a_2{}^2-b_2{}^2}=\sqrt{81-9}=\sqrt{72}=\pm 6\sqrt{2}\end{array}$

Therefore, the eccentricity for first graph is $\frac{c_1}{a_1}=\frac{3\sqrt{5}}{9}=\frac{\sqrt{5}}{3}$ .

And the eccentricity for second graph is $\frac{c_2}{a_2}=\frac{6\sqrt{2}}{9}=\frac{2\sqrt{2}}{3}$ .

d.

To determine

To make the conjecture about the eccentricity and shape of each ellipse.

Explanation of Solution

Given:

The given equations of ellipses are -

  $\frac{x^2}{81}+\frac{y^2}{36}=1$ And $\frac{x^2}{81}+\frac{y^2}{9}=1$

Calculation:

As the eccentricity of an ellipse approaches zero, it approaches a circle. The larger the eccentricity, the more stretched the ellipse becomes, and the more it deviates from a pure circle in shape.