Chapter 9.2, Problem 21E

To determine

To find:The standard equation of the ellipse with the given characteristics in the graph.

  

Answer to Problem 21E

The equation of the ellipse is $\frac{{\left(x-2\right)}^2}{2^2}+\frac{{\left(y+1\right)}^2}{1^2}=1$ .

Explanation of Solution

Given information:

The graph of the ellipse is shown below.

  

Calculation:

The major axis of the ellipse is horizontal. So, the general equation of ellipse is $\frac{{\left(x-h\right)}^2}{a^2}+\frac{{\left(y-k\right)}^2}{b^2}=1$ .

From the graph, the vertices are $\left(0,-1\right),\left(4,-1\right)$ and $\left(2,0\right),\left(2,-2\right)$ .

Therefore, the center of the ellipse is at midpoint of the vertices.

  $\frac{1}{2}\left[\left(0,-1\right)+\left(4,-1\right)\right]=\left(2,-1\right)$

So, the center of the ellipse is the point $\left(2,-1\right)$ .

The length of the major axis is the distance between the points $\left(0,-1\right)$ and $\left(4,-1\right)$ .

  $\begin{array}{c}\sqrt{{\left(0-4\right)}^2+{\left(-1-\left(-1\right)\right)}^2}=\sqrt{4^2}\\ =4\end{array}$

The length of the minor axis is the distance between the points $\left(2,0\right)$ and $\left(2,-2\right)$ .

  $\begin{array}{c}\sqrt{{\left(2-2\right)}^2+{\left(0-\left(-2\right)\right)}^2}=\sqrt{2^2}\\ =2\end{array}$

Use the formula of length of major axis.

  $\begin{array}{c}2a=4\\ a=2\end{array}$

Use the formula of length of minor axis.

  $\begin{array}{c}2b=2\\ b=1\end{array}$

Substitute $2$ for $a$ , $1$ for $b$ , $2$ for $h$ and $-1$ for $k$ in the general equation of ellipse.

  $\begin{array}{c}\frac{{\left(x-h\right)}^2}{a^2}+\frac{{\left(y-k\right)}^2}{b^2}=1\\ \frac{{\left(x-2\right)}^2}{2^2}+\frac{{\left(y-\left(-1\right)\right)}^2}{1^2}=1\\ \frac{{\left(x-2\right)}^2}{2^2}+\frac{{\left(y+1\right)}^2}{1^2}=1\end{array}$

Therefore, the equation of the ellipse is $\frac{{\left(x-2\right)}^2}{2^2}+\frac{{\left(y+1\right)}^2}{1^2}=1$ .