Chapter 9.4, Problem 36PPS

To determine

To write: The equation for the given ellipse.

Answer to Problem 36PPS

The equation for the given ellipse is $\frac{{\left(x-4\right)}^2}{16}+\frac{{\left(y-9\right)}^2}{41}=1$ .

Explanation of Solution

Given information:

The foci of the ellipse are at $\left(4,4\right),\left(4,14\right)$ and the co-vertices is at $\left(0,9\right)$ .

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 $k$ is calculated as,

  $\begin{array}{l}\left(h,k\right)=\left(\frac{4+4}{2},\frac{4+14}{2}\right)\\ \left(h,k\right)=\left(4,9\right)\end{array}$

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

  $\begin{array}{l}b=4-0\\ b=4\end{array}$

The value of $c$ is calculated as,

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

The value of $a$ is calculated as,

  $\begin{array}{l}c=\sqrt{a^2-b^2}\\ 5=\sqrt{a^2-4^2}\\ a=\sqrt{41}\end{array}$

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

  $\begin{array}{l}\frac{{\left(x-h\right)}^2}{b^2}+\frac{{\left(y-k\right)}^2}{a^2}=1\\ \frac{{\left(x-4\right)}^2}{{\left(4\right)}^2}+\frac{{\left(y-9\right)}^2}{{\left(\sqrt{41}\right)}^2}=1\\ \frac{{\left(x-4\right)}^2}{16}+\frac{{\left(y-9\right)}^2}{41}=1\end{array}$

Thus, the equation for the given ellipse is $\frac{{\left(x-4\right)}^2}{16}+\frac{{\left(y-9\right)}^2}{41}=1$ .