Chapter 11.2, Problem 20E

To determine

To determine the vertices, foci, eccentricity, length of major and minor axes of the ellipse and sketch the graph.

Answer to Problem 20E

  $\begin{array}{l}\text{Vertices}=\left(\pm \text{a,0}\right)=\left(\pm 2,0\right)\\ \text{Foci}=\left(\pm \text{c,0}\right)=\left(\pm \sqrt{2},0\right)\\ \text{Eccentricity,e}=\frac{\text{c}}{\text{a}}=\frac{\sqrt{2}}{2}\\ \text{The length of major axis}=\text{2a}=4\\ \text{The length of minor axis}=\text{2b}=2\sqrt{2}\end{array}$

  

Explanation of Solution

Given information:

The given equation is:

  ${\text{x}}^{\text{2}}=4-2{\text{y}}^{\text{2}}$

Formula used:

The following formulas are used:

  $\begin{array}{l}{\text{c}}^2={\text{a}}^{\text{2}}-{\text{b}}^2\\ \text{Vertices}=\left(\pm \text{a,0}\right)\\ \text{Foci}=\left(\pm \text{c,0}\right)\\ \text{Eccentricity,e}=\frac{\text{c}}{\text{a}}\\ \text{The length of major axis}=\text{2a}\\ \text{The length of minor axis}=\text{2b}\end{array}$

Calculation:

In the equation ${\text{x}}^{\text{2}}=4-2{\text{y}}^{\text{2}}{\text{ or x}}^2+2{\text{y}}^2=4\text{ or }\frac{{\text{x}}^2}{4}+\frac{{\text{y}}^2}{2}=1$

The denominator of ${\text{x}}^{\text{2}}$ is greater, so the ellipse has a major horizontal axis.

  $\begin{array}{l}{\text{a}}^{\text{2}}=4\text{ or a}=\pm 2\\ {\text{b}}^2=2\text{ or b}=\pm \sqrt{2}\\ {\text{c}}^2={\text{a}}^{\text{2}}-{\text{b}}^2=4-2=2\text{ or c}=\pm \sqrt{2}\\ \text{Vertices}=\left(\pm \text{a,0}\right)=\left(\pm 2,0\right)\\ \text{Foci}=\left(\pm \text{c,0}\right)=\left(\pm \sqrt{2},0\right)\\ \text{Eccentricity,e}=\frac{\text{c}}{\text{a}}=\frac{\sqrt{2}}{2}\\ \text{The length of major axis}=\text{2a}=4\\ \text{The length of minor axis}=\text{2b}=2\sqrt{2}\end{array}$

To sketch the graph, solve for $\text{y}$ :

  $\begin{array}{l}\frac{{\text{x}}^{\text{2}}}{4}+\frac{{\text{y}}^{\text{2}}}{2}=1\\ \frac{{\text{y}}^{\text{2}}}{2}=1-\frac{{\text{x}}^{\text{2}}}{4}\\ {\text{y}}^{\text{2}}=2\left(1-\frac{{\text{x}}^{\text{2}}}{4}\right)\\ \text{y}=\pm \sqrt{2\left(1-\frac{{\text{x}}^{\text{2}}}{4}\right)}\\ \text{y}=\sqrt{2\left(1-\frac{{\text{x}}^{\text{2}}}{4}\right)}\text{ or }-\sqrt{2\left(1-\frac{{\text{x}}^{\text{2}}}{4}\right)}\end{array}$

The graph is given below: