Chapter 9.2, Problem 53E

To determine

To find: The equation of the ellipse with foci $\left(1,1\right),\left(1,13\right)$ and eccentricity $\frac{2}{3}$ .

Answer to Problem 53E

The equation of ellipse is $\frac{{\left(x-1\right)}^2}{45}+\frac{{\left(y-5\right)}^2}{81}=1$ .

Explanation of Solution

Given information:

The foci of the ellipse are $\left(1,1\right),\left(1,13\right)$ and eccentricity is $\frac{2}{3}$ .

Calculation:

The foci of ellipse lies on the y-axis. So, The major axis of the ellipse is vertical.

So, the general equation of ellipse is $\frac{{\left(x-h\right)}^2}{b^2}+\frac{{\left(y-k\right)}^2}{a^2}=1$ .

As the foci of the ellipse is $\left(h,k\pm c\right)=\left(1,5\pm 6\right)$ .

Therefore, the value of $h=1$ and $k=5$ .

Use the formula for eccentricity.

  $e=\frac{c}{a}$

So, The value of $c=6$ and $a=9$ as the foci are $\left(1,1\right),\left(1,13\right)$ .

Use the formula.

  $\begin{array}{c}{c}^2=a^2-b^2\\ 6^2=9^2-b^2\\ b^2=81-36\\ b=3\sqrt{5}\\ b\approx 6.71\end{array}$

Now substitute $9$ for $a$ and $3\sqrt{5}$ for $b$ in the general equation of ellipse.

  $\begin{array}{c}\frac{{\left(x-h\right)}^2}{b^2}+\frac{{\left(y-k\right)}^2}{a^2}=1\\ \frac{{\left(x-1\right)}^2}{{\left(3\sqrt{5}\right)}^2}+\frac{{\left(y-5\right)}^2}{9^2}=1\\ \frac{{\left(x-1\right)}^2}{45}+\frac{{\left(y-5\right)}^2}{81}=1\end{array}$

Therefore, the equation of ellipse is $\frac{{\left(x-1\right)}^2}{45}+\frac{{\left(y-5\right)}^2}{81}=1$ .