Chapter 12.7, Problem 60SR

To determine

To find: Equation of ellipse.

Answer to Problem 60SR

  $\frac{{\left(x-2\right)}^2}{9}+\frac{{\left(y-7\right)}^2}{45}=1$

Explanation of Solution

Given information:

Foci at $\left(2,1\right)$ and $\left(2,13\right)$

Co-vertex $\left(5,7\right)$

Given

Foci at $\left(2,1\right)$ and $\left(2,13\right)$

On comparing the both foci we can say that x-coordinate is the same, so the ellipse has vertical orientation.

Therefore,

Standard form of ellipse is

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

Center $\left(h,k\right)$

Foci $\left(h,k\pm c\right)$

Center of ellipse is the mid point between the foci, so

Therefore,

Center $=\left(\frac{2+2}{2},\frac{1+13}{2}\right)$

  $\Rightarrow$ Center $=\left(\frac{4}{2},\frac{14}{2}\right)$

  $\Rightarrow$ Center $=\left(2,7\right)$

On comparing

We get

  $\begin{array}{l}h=2\\ k=7\end{array}$

Distance between Foci and center will be $c$

Therefore,

Using distance formula between $\left(2,7\right)$ and $\left(2,13\right)$

  $\begin{array}{l}\Rightarrow c=\sqrt{{\left(2-2\right)}^2+{\left(13-7\right)}^2}\\ \Rightarrow c=\sqrt{{\left(0\right)}^2+{\left(6\right)}^2}\\ \Rightarrow c=\sqrt{{\left(6\right)}^2}\\ \Rightarrow c=6\end{array}$

We know distance between center $\left(2,7\right)$ and co-vertex $\left(5,7\right)$ will be $b$

Therefore,

We get

  $\begin{array}{l}\Rightarrow b=\sqrt{{\left(5-2\right)}^2+{\left(7-7\right)}^2}\\ \Rightarrow b=\sqrt{{\left(3\right)}^2+{\left(0\right)}^2}\\ \Rightarrow b=\sqrt{{\left(3\right)}^2}\\ \Rightarrow b=3\end{array}$

We know

  $c^2=a^2-b^2$

On substituting values

We get

  $\begin{array}{l}{c}^2=a^2-b^2\\ \Rightarrow 6^2=a^2-3^2\\ \Rightarrow 36=a^2-9\\ \Rightarrow a^2=36+9\\ \Rightarrow a^2=45\\ \Rightarrow a=\sqrt{45}\end{array}$

Therefore,

Required equation of ellipse is

  $\begin{array}{l}\frac{{\left(x-h\right)}^2}{b^2}+\frac{{\left(y-k\right)}^2}{a^2}=1\\ \Rightarrow \frac{{\left(x-2\right)}^2}{3^2}+\frac{{\left(y-7\right)}^2}{{\left(\sqrt{45}\right)}^2}=1\\ \Rightarrow \frac{{\left(x-2\right)}^2}{9}+\frac{{\left(y-7\right)}^2}{45}=1\end{array}$