Chapter 11.2, Problem 48E

a.

To determine

To determine the equation of the ancillary circle.

Answer to Problem 48E

The equation of the ancillary circle is ${\text{x}}^2+{\text{y}}^2=4$

Explanation of Solution

Given information:

The radius of the ancillary circle is equal to half the length of the minor axis and center is the same as the ellipse. The ancillary circle is the largest circle that can fit inside an ellipse.

The equation of the ellipse is ${\text{x}}^{\text{2}}+4{\text{y}}^2=16$

Calculation:

The value of $\text{a}$ and $\text{b}$ can be found from the equation of the ellipse:

  $\begin{array}{l}{\text{x}}^{\text{2}}+4{\text{y}}^2=16\\ \frac{{\text{x}}^{\text{2}}}{16}+\frac{{\text{y}}^2}{4}=1\\ {\text{a}}^2=16\\ \text{a}=\pm 4\\ {\text{b}}^2=4\\ \text{b}=\pm 2\end{array}$

Radius of the circle is half the length of the minor axis:

  $\begin{array}{l}\text{Length of the minor axis}=\text{2b}\\ \text{2b}=4\end{array}$

The radius of the circle is $\text{b}=2$

The equation of the circle can be found by:

  $\begin{array}{l}{\text{x}}^2+{\text{y}}^2=2^2\\ {\text{x}}^2+{\text{y}}^2=4\end{array}$

Hence, the equation of the ancillary circle is ${\text{x}}^2+{\text{y}}^2=4$

b.

To determine

To show that for a point $\left(\text{s,t}\right)$ on the ancillary circle, $\left(\text{2s,t}\right)$ is the point on the ellipse.

Explanation of Solution

Given information:

The equation of the ellipse is ${\text{x}}^{\text{2}}+4{\text{y}}^2=16\text{ or }\frac{{\text{x}}^{\text{2}}}{16}+\frac{{\text{y}}^2}{4}=1$

The equation of the ancillary circle is ${\text{x}}^2+{\text{y}}^2=4$

Proof:

The equation of the circle can be written by using the coordinates $\left(\text{s,t}\right)$ :

  $\begin{array}{l}{\text{x}}^2+{\text{y}}^2=4\\ {\text{s}}^2+{\text{t}}^2=4\left[\text{Substituting x and y}\right]\\ \frac{{\text{s}}^2}{4}+\frac{{\text{t}}^2}{4}=1\end{array}$

The equation of the ellipse can be written by using the coordinates $\left(\text{s,t}\right)$ :

  $\begin{array}{l}{\text{x}}^{\text{2}}+4{\text{y}}^2=16\\ \frac{{\text{x}}^{\text{2}}}{16}+\frac{{\text{y}}^2}{4}=1\\ \frac{{\text{s}}^2}{16}+\frac{{\text{t}}^2}{4}=1\left[\text{Substituting x and y}\right]\\ \frac{{\text{4s}}^2}{4}+\frac{{\text{t}}^2}{4}=1\\ \frac{{\text{2}}^2{\text{s}}^2}{4}+\frac{{\text{t}}^2}{4}=1\\ \frac{{\left(\text{2s}\right)}^2}{4}+\frac{{\text{t}}^2}{4}=1\end{array}$

Comparing the equations $\frac{{\text{s}}^2}{4}+\frac{{\text{t}}^2}{4}=1$ and $\frac{{\left(\text{2s}\right)}^2}{4}+\frac{{\text{t}}^2}{4}=1$ , it is seen that for a point $\left(\text{s,t}\right)$ on the ancillary circle, $\left(\text{2s,t}\right)$ is the point on the ellipse.