Chapter 8.7, Problem 8P

Solution

To write: the equation of the ellipse.

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

Given information:

The foci of the ellipse are at $\left(-4,2\right)$ and $\left(-4,-2\right)$ and $e=0.5$ .

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 $c^2=a^2-b^2$

Eccentricity of conic sections

The eccentricity of each conic section is defined below. For an ellipse or hyperbola, c is the distance from each focus to the center, and a is the distance from each vertex to center.

Circle: $e=0$

Parabola: $e=1$

Ellipse: $e=\frac{c}{a},\text{ and }0<e<1$

Hyperbola: $e=\frac{c}{a},\text{ and }e>1$

Explanation:

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

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

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

The center is between -2 and 2.

  $\begin{array}{c}k=\frac{-2+2}{2}\\ =\frac{0}{2}\\ =0\end{array}$

So, the center is at $\left(h,k\right)=\left(-4,0\right)$ .

The foci of ellipse is at $\left(-4,2\right)$ and $\left(-4,-2\right)$ . Distance from center to focus is:

  $\begin{array}{c}c=2-k\\ =2-0\\ =2\end{array}$

It is given that the eccentricity of the given ellipse is:

  $\begin{array}{c}e=0.5\\ \frac{c}{a}=\frac{5}{10}\\ \frac{2}{a}=\frac{1}{2}\\ a=4\end{array}$

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.

  $\begin{array}{c}\frac{{\left(x-h\right)}^2}{b^2}+\frac{{\left(y-k\right)}^2}{a^2}=1\\ \frac{{\left(x-\left(-4\right)\right)}^2}{12}+\frac{{\left(y-0\right)}^2}{4^2}=1\\ \frac{{\left(x+4\right)}^2}{12}+\frac{y^2}{16}=1\end{array}$