Chapter 10.3, Problem 34E

To determine

Find the center, vertices, foci and eccentricity of the ellipse and sketch the ellipse

Answer to Problem 34E

Center: $\left(0,0\right)$

Vertices: $\left(0,9\right),\left(0,-9\right)$

Foci: $\left(0,\sqrt{65}\right),\left(0,-\sqrt{65}\right)$

Eccentricity: $e=\frac{\sqrt{65}}{9}$

Explanation of Solution

Given:

The given equation is

  $\frac{x^2}{16}+\frac{y^2}{81}=1$

  $\frac{x^2}{16}+\frac{y^2}{81}=\mathrm{1.............}(1)$

The standard equation of an ellipse with major axis vertical is given by

  $\frac{{(x-h)}^2}{b^2}+\frac{{(y-k)}^2}{a^2}=1$

Comparing this with equation (1),

  $h=0\text{ and}k=0$

Thus, the center of the ellipse is (0, 0).

  $a=9,b=4$

Now, since the vertices are at $(0,\pm a)$ , the two vertices to be $(0,9),(0,-9)$

Next, we have

  $\begin{array}{l}{c}^2=a^2-b^2\\ =81-16\\ =65\end{array}$

Thus, $c=\sqrt{65}$

The foci lie on the points $(h,k\pm c)$

The foci are at $(0,\sqrt{65}),(0,-\sqrt{65})$

The eccentricity of the ellipse is given by

  $e=\frac{c}{a}$

Thus,

  $e=\frac{\sqrt{65}}{9}$

The graph of the given ellipse is depicted below: