Chapter 8.7, Problem 7P

Solution

To write: the equation of the ellipse.

  $\frac{x^2}{36}+\frac{25{\left(y-6\right)}^2}{756}=1$

Given information:

The vertices of the ellipse are at $\left(-6,4\right)$ and $\left(6,4\right)$ and $e=0.4$ .

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 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 -6 and 6.

  $\begin{array}{c}h=\frac{-6+6}{2}\\ =\frac{0}{2}\\ =0\end{array}$

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

Distance from center to vertex a :

  $\begin{array}{c}a=6-h\\ =6-0\\ =6\end{array}$

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

  $\begin{array}{c}e=0.4\\ \frac{c}{a}=\frac{4}{10}\\ \frac{c}{6}=\frac{2}{5}\\ c=\frac{12}{5}\end{array}$

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

  $\begin{array}{c}{c}^2=a^2-b^2\\ {\left(\frac{12}{5}\right)}^2=6^2-b^2\\ \frac{144}{25}=36-b^2\\ b^2=36-\frac{144}{25}\\ b^2=\frac{900-144}{25}\\ b^2=\frac{756}{25}\end{array}$

Now susbtitute these values in the standard equation of ellipse.

  $\begin{array}{c}\frac{{\left(x-h\right)}^2}{a^2}+\frac{{\left(y-k\right)}^2}{b^2}=1\\ \frac{{\left(x-0\right)}^2}{6^2}+\frac{{\left(y-6\right)}^2}{\frac{756}{25}}=1\\ \frac{x^2}{36}+\frac{25{\left(y-6\right)}^2}{756}=1\end{array}$