Chapter 10.3, Problem 36E

To determine

To write: The equation of the ellipse whose foci are at $\left(1,-1\right)\text{and}\left(1,5\right)$ and the ellipse passes through the point at $\left(4,2\right)$ .

Answer to Problem 36E

  $\frac{{\left(x-1\right)}^2}{9}+\frac{{\left(y-2\right)}^2}{18}=1$

Explanation of Solution

Given information: Foci are at $\left(1,-1\right)\text{and}\left(1,5\right)$ and the ellipse passes through the point at $\left(4,2\right)$

Calculation:

We know from the foci, that the center is at: $\left(1,2\right)$ and that the major axis is vertical, giving us:

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

We can also use the foci points to find c = 3. From this we know

  $a^2-b^2=c^2\to b^2=a^2-9$

Substitute this for $b^2$ and the point $\left(4,2\right)$ for x , y and solve for $a^2$ .

  $\begin{array}{l}\frac{{\left(4-1\right)}^2}{a^2-9}+\frac{{\left(2-2\right)}^2}{a^2}=1\\ a^2-9=3^2\to a^2=18\\ \Rightarrow b^2=18-9=9\end{array}$

This gives us the final equation:

  $\frac{{\left(x-1\right)}^2}{9}+\frac{{\left(y-2\right)}^2}{18}=1$