Chapter EP, Problem 10.4.1EP

To determine

To find: The equation of ellipse that satisfies each set of conditions.

Answer to Problem 10.4.1EP

The equation of ellipse is given below.

Explanation of Solution

Given:

The end points of major axisvertices are at $\left(-2,7\right)\text{and}\left(4,7\right)$ .

The end points of minor axis co-vertices are at $\left(1,5\right)\text{and}\left(1,9\right)$ .

The standard form of the equation of ellipse that is horizontal with centre $\left(h,k\right)$ and length of major axis is $2a$ and length of minor axis is $2b$ , is given as:

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

As the $y$ -coordinate is the same for both vertices, so the ellipse is horizontal.

The centre of the ellipse is calculated as:

  $\begin{array}{c}\left(h,k\right)=\left(\frac{-2+4}{2},\frac{7+7}{2}\right)\\ =\left(1,7\right)\end{array}$

The length of the major axis $2a$ is given as:

  $4-\left(-2\right)=6$

So the value of $a$ is calculated as:

  $\begin{array}{c}2a=6\\ a=3\end{array}$

The length of the minor axis $2b$ is given as:

  $9-5=4$

So the value of $b$ is calculated as:

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

So the equation for the ellipse is given as:

  $\frac{{\left(x-1\right)}^2}{9}+\frac{{\left(y-7\right)}^2}{4}=1\left[\because a^2=9,b^2=4\right]$