Chapter 9.2, Problem 8E

To determine

To check:The ellipse $\frac{x^2}{2^2}+\frac{y^2}{8^2}=1$ is elongated or nearly circular.

Answer to Problem 8E

The ellipse $\frac{x^2}{2^2}+\frac{y^2}{8^2}=1$ is elongated.

Explanation of Solution

Given information:

The equation of the ellipse is $\frac{x^2}{2^2}+\frac{y^2}{8^2}=1$ .

Formula used:

The eccentricity of ellipse is $e=\frac{c}{a}$ .

Calculation:

Compare the given equation of ellipse with general equation of ellipse $\frac{x^2}{b^2}+\frac{y^2}{a^2}=1$ .

  $b=2$ and $a=8$ .

Calculate the value of $c$ .

  $\begin{array}{c}c=\sqrt{a^2-b^2}\\ =\sqrt{8^2-2^2}\\ =\sqrt{64-4}\\ =\sqrt{60}\\ \approx 7.75\end{array}$

Calculate the eccentricity of the ellipse.

  $\begin{array}{c}e=\frac{c}{a}\\ =\frac{7.75}{8}\\ \approx 0.97\end{array}$

Since the eccentricity of the ellipse is nearly equal to $1$ . So, the ellipse is elongated.

Therefore, the ellipse $\frac{x^2}{2^2}+\frac{y^2}{8^2}=1$ is elongated.