Chapter 9.2, Problem 29E

To determine

To find:The center, vertices, foci and eccentricity of the ellipse $\frac{x^2}{64}+\frac{y^2}{9}=1$ , sketch its graph. Verify the result graphically.

Answer to Problem 29E

The center, vertices, foci and eccentricity of the ellipse $\frac{x^2}{64}+\frac{y^2}{9}=1$ are $\left(0,0\right)$ , $\left(\pm 8,0\right)$ , $\left(\pm \sqrt{55},0\right)$ and $0.93$ respectively. The graph of the ellipse is shown in the Figure(1).

Explanation of Solution

Given information:

The equation of the ellipse is $\frac{x^2}{64}+\frac{y^2}{9}=1$ .

Calculation:

Compare the equation of ellipse $\frac{x^2}{64}+\frac{y^2}{9}=1$ with general equation $\frac{{\left(x-h\right)}^2}{a^2}+\frac{{\left(y-k\right)}^2}{b^2}=1$ .

It gives $a=8$ , $b=3$ , $h=0$ and $k=0$ . The major axis of the ellipse is horizontal.

Use the formula.

  $\begin{array}{c}{c}^2=a^2-b^2\\ c^2=8^2-3^2\\ c^2=64-9\\ c^2=55\\ c=\sqrt{55}\end{array}$

Use the formula for eccentricity.

  $\begin{array}{c}e=\frac{c}{a}\\ =\frac{\sqrt{55}}{8}\\ \approx 0.93\end{array}$

The center of the ellipse is $\left(h,k\right)=\left(0,0\right)$ , vertices of the ellipse is $\left(h\pm a,k\right)=\left(\pm 8,0\right)$ , foci of the ellipse is $\left(h\pm c,k\right)=\left(\pm \sqrt{55},0\right)$ and eccentricity of ellipse is $e=0.93$ .

Therefore, The center, vertices, foci and eccentricity of the ellipse $\frac{x^2}{64}+\frac{y^2}{9}=1$ are $\left(0,0\right)$ , $\left(\pm 8,0\right)$ , $\left(\pm \sqrt{55},0\right)$ and $0.93$ respectively.

The graph of the ellipse $\frac{x^2}{64}+\frac{y^2}{9}=1$ is shown below.

  

Figure(1)

Verify the result graphically:

The graph of the ellipse using graphing utility is shown below.

  

Figure(2)

From the above graphs, we can see that both the graphs show same value at each point.

Therefore, both the graphs are same.