Chapter EP, Problem 10.4.11EP

To determine

To find: The coordinates of the centre and foci and the length of major and minor axes for the ellipse with the given equation. And hence graph the ellipse.

Answer to Problem 10.4.11EP

The coordinates of the centre and foci and the length of major and minor axes for the ellipse are given below.

Explanation of Solution

Given:

The given equation of the ellipse is as:

  $x^2+4y^2+8x-64y=-128$

Rewrite the given equation as:

  $\begin{array}{l}{x}^2+8x+4y^2-64y=-128\\ x^2+8x+16-16+4\left(y^2-16y+64-64\right)=-128\\ {\left(x+4\right)}^2+4{\left(y-8\right)}^2=144\\ \frac{{\left(x+4\right)}^2}{144}+\frac{{\left(y-8\right)}^2}{36}=1\end{array}$

Consider the standard equation of the ellipse, as:

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

By comparing the given equation of ellipse with the standard equation of the ellipse,

The centre is given as:

  $\left(h,k\right)=\left(-4,8\right)$ .

The value of $a^2$ and $b^2$ is as:

  $\begin{array}{c}{b}^2=36\\ b=6\\ a^2=144\\ a=12\end{array}$

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

  $\begin{array}{c}2a=2\left(12\right)\\ 2a=24\end{array}$

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

  $\begin{array}{c}2b=2\left(6\right)\\ 2b=12\end{array}$

The focal length $c$ is calculated as:

  $\begin{array}{c}c=\sqrt{a^2-b^2}\\ =\sqrt{144-36}\\ =\sqrt{108}\\ =10.392\end{array}$

The coordinate of foci is given as:

  $\left(h\pm c,k\right)=\left(-4\pm 10.392,8\right)$ .

So the coordinate of the foci are as:

  $\left(6.392,8\right),\left(-14.392,8\right)$ .

The graph of the ellipse is given below: