Chapter 8.6, Problem 17E

Solution

To write: the equation of the ellipse.

  $\frac{{\left(x-1\right)}^2}{16}+\frac{{\left(y-4\right)}^2}{12}=1$

Given information:

The vertices of the ellipse are at $\left(-3,4\right)$ and $\left(5,4\right)$ and foci at $\left(-1,4\right)$ and $\left(3,4\right)$ .

Formula Used:

Standard equation of an Ellipse

Equation	Major Axis	Vertices	Co-vertices
$\frac{{\left(x-h\right)}^2}{a^2}+\frac{{\left(y-k\right)}^2}{b^2}=1$	Horizontal	$\left(h\pm a,k\right)$	$\left(h,k\pm b\right)$
$\frac{{\left(x-h\right)}^2}{b^2}+\frac{{\left(y-k\right)}^2}{a^2}=1$	Vertical	$\left(h,k\pm a\right)$	$\left(h\pm b,k\right)$

The center of the Ellipse is $\left(h,k\right)$ . The major and minor axes are of the lengths $2a$ and $2b$ respectively, where $a>b>0$ . The foci of the ellipse lie on the major axis at a distance of c units from the center, where $c^2=a^2-b^2$

Explanation:

Center of the ellipse is at the middle of the two vertices.

Since, the y co-ordinates are same, so it has horizontal major axis and has the form:

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

The center is between -3 and 5.

  $\begin{array}{c}h=\frac{-3+5}{2}\\ =\frac{2}{2}\\ =1\end{array}$

So, the center is at $\left(h,k\right)=\left(1,4\right)$ .

Distance from center to vertex a :

  $\begin{array}{c}a=5-h\\ =5-1\\ =4\end{array}$

The foci of ellipse is at $\left(-1,4\right)$ and $\left(3,4\right)$ . Distance from center to focus is:

  $\begin{array}{c}c=3-h\\ =3-1\\ =2\end{array}$

Now, find the distance from the center to co-vertex b .

  $\begin{array}{c}{c}^2=a^2-b^2\\ 2^2=4^2-b^2\\ 4=16-b^2\\ b^2=16-4\\ b^2=12\end{array}$

Now susbtitute these values in the standard equation of ellipse.

  $\frac{{\left(x-1\right)}^2}{16}+\frac{{\left(y-4\right)}^2}{12}=1$