Chapter 9.2, Problem 38E

To determine

To find the center, vertices, foci and eccentricity of the ellipse and graph the ellipse.

Answer to Problem 38E

The center of the ellipse is $\left(0,0\right)$ .

  $\begin{array}{l}\text{Vertices}=\left(\pm a,0\right)=\left(0,\pm 4\right)\\ \text{Foci}=\left(\pm c,0\right)=\left(0,\pm \sqrt{15}\right)\\ \text{Eccentricity,}e=\frac{c}{a}=\frac{\sqrt{15}}{4}\end{array}$

Explanation of Solution

Given information:

The equation of the ellipse is:

  $16x^2+y^2=16$

Formula used:

The following formula is used:

  $\begin{array}{l}\text{Vertices}=\left(\pm a,0\right)\\ \text{Foci}=\left(\pm c,0\right)\\ \text{Eccentricity,}e=\frac{c}{a}\end{array}$

Calculation:

Change the equation to standard form:

  $\begin{array}{l}16x^2+y^2=16\\ \frac{16x^2+y^2}{16}=1\\ \frac{16x^2}{16}+\frac{y^2}{16}=1\\ \frac{x^2}{1}+\frac{y^2}{16}=1\end{array}$

The ellipse has a major vertical axis.

The center of the ellipse is $\left(0,0\right)$ .

  $\begin{array}{l}{a}^2\text{=16 or }a=\pm 4\\ b^2=1\text{ or }b=\pm 1\\ c^2=a^2-b^2=16-1=15\text{ or }c=\pm \sqrt{15}\\ \text{Vertices}=\left(\pm a,0\right)=\left(0,\pm 4\right)\\ \text{Foci}=\left(\pm c,0\right)=\left(0,\pm \sqrt{15}\right)\\ \text{Eccentricity,}e=\frac{c}{a}=\frac{\sqrt{15}}{4}\end{array}$

To graph the ellipse solve for $y$ :

  $\begin{array}{l}\frac{x^2}{1}+\frac{y^2}{16}=1\\ \frac{y^2}{16}=1-x^2\\ y^2=\pm \sqrt{16\left(1-x^2\right)}\\ y=\pm 4\sqrt{\left(1-x^2\right)}\\ y=4\sqrt{\left(1-x^2\right)}\text{ or }-4\sqrt{\left(1-x^2\right)}\end{array}$

The graph is given below: