Chapter 9.4, Problem 34PPS

To determine

To write: The equation for the given ellipse.

Answer to Problem 34PPS

The equation for the given ellipse is $\frac{{\left(x-4\right)}^2}{29}+\frac{{\left(y+3\right)}^2}{4}=1$ .

Explanation of Solution

Given information:

The focus of the ellipse is at $\left(9,-3\right)$ , the co-vertices is at $\left(4,-5\right)$ , and the centre is at $\left(4,-3\right)$ .

Concept used:

Write the standard form of the equation of the ellipse whose orientation is along the horizontal axis is,

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

The coordinate of the foci is $\left(h\pm c,k\right)$

The value of $c$ is calculated as,

  $c=\sqrt{a^2-b^2}$

The vertices are at $\left(h\pm a,k\right)$ and the co-vertices are at $\left(h,k\pm b\right)$ .

Write the standard form of the equation of the ellipse whose orientation is along the vertical axis is,

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

The coordinate of the foci is $\left(h,k\pm c\right)$

The vertices are at $\left(h,k\pm a\right)$ and the co-vertices are at $\left(h\pm b,k\right)$ .

The value of $c$ is calculated as,

  $c=\sqrt{a^2-b^2}$

The value of $h$ and $k$ from the centre is,

  $\begin{array}{l}h=4\\ k=-3\end{array}$

The value of $b$ from the covertex and the focus is calculated as,

  $\begin{array}{l}b=\left(-3\right)-\left(-5\right)\\ b=2\end{array}$

The value of $c$ is calculated as,

  $\begin{array}{c}c=9-4\\ =5\end{array}$

The value of $a$ is calculated as,

  $\begin{array}{l}c=\sqrt{a^2-b^2}\\ 5=\sqrt{a^2-2^2}\\ a=\sqrt{29}\end{array}$

The standard form of the equation of the given ellipse is,

  $\begin{array}{l}\frac{{\left(x-h\right)}^2}{a^2}+\frac{{\left(y-k\right)}^2}{b^2}=1\\ \frac{{\left(x-4\right)}^2}{{\left(\sqrt{29}\right)}^2}+\frac{{\left(y-\left(-3\right)\right)}^2}{{\left(2\right)}^2}=1\\ \frac{{\left(x-4\right)}^2}{29}+\frac{{\left(y+3\right)}^2}{4}=1\end{array}$

Thus, the equation for the given ellipse is $\frac{{\left(x-4\right)}^2}{29}+\frac{{\left(y+3\right)}^2}{4}=1$ .