Chapter 9.4, Problem 20MCQ

To determine

To find: find the equation of the ellipse.

Answer to Problem 20MCQ

The equation of an ellipse is $\frac{{\left(y-2\right)}^2}{64}+\frac{{\left(x+4\right)}^2}{36}=1$

Explanation of Solution

Given information:

The given endpoints of ellipse is,

  $\left(-4,10\right),\left(-4,-6\right)$

And foci

  $\left(-4,7.3\right)$ , $\left(-4,3.3\right)$

The midpoint of an ellipse is the centre of the circle is calculated as,

The given endpoints are

  $\left(x_1=-4,y_1=10\right)$ and $\left(x_2=-4,y_2=-6\right)$

Simplify,

  $\begin{array}{l}M=\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\\ M=\frac{-4+\left(-4\right)}{2},\frac{10+\left(-6\right)}{2}\\ M=-4,2\end{array}$

The centre of the ellipse is $\left(-4,2\right)$

The length of the major axis is

  $\begin{array}{l}2a=10-\left(-6\right)\\ a=8\end{array}$

The distance of the foci is

Foci= $2c$

  $\begin{array}{l}2c=7.3-\left(-3.3\right)\\ c=5.3\end{array}$

The relation of between $a,b$ and c are

  $\begin{array}{l}{a}^2-b^2=c^2\\ {\left(8\right)}^2-b^2={\left(5.3\right)}^2\\ b^2=64-28\\ b=6\end{array}$

By using the equation of an ellipse with vertical major axis,

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

Put the values of $h=-4,k=2$ and $a=8,b=6$ in the equation (2),

  $\begin{array}{l}\frac{{\left(y-k\right)}^2}{a^2}+\frac{{\left(x-h\right)}^2}{b^2}=1\\ \frac{{\left(y-2\right)}^2}{8^2}+\frac{{\left(x-\left(-4\right)\right)}^2}{6^2}=1\\ \frac{{\left(y-2\right)}^2}{64}+\frac{{\left(x+4\right)}^2}{36}=1\end{array}$

Thus, the equation (H) represents an ellipse.