Question

How do i solve this for standard equation and foci?

Answer 1

For this problem, we are given the graph of an ellipse, and we need to determine its expression in the standard form.

The standard equation of an ellipse is given below:

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

Where (h,k) is the center of the ellipse, a is the horizontal radius and b is the vertical radius.

The center of the ellipse on our problem is (-2,2), the vertical radius is 2 and the horizontal radius is 3. We have:

$\begin{gathered} \frac{(x+2)^2}{3^2}+\frac{(y-2)^2}{2^2}=1 \\ \frac{(x+2)^2}{9}+\frac{(y-2)^2}{4}=1 \end{gathered}$

In order to calculate the Foci, we need to first find the eccentricity of the ellipse, which is given by the following formula:

$\begin{gathered} e=\sqrt{a^2-b^2} \\ e=\sqrt{3^2-2^2} \\ e=\sqrt{9-4}=\sqrt{5} \end{gathered}$

The coordinates of the foci are given by:

$\begin{gathered} F(h+e,k)=(-2-\sqrt{5},2) \\ F^{\prime}(h-e,k)=(-2+\sqrt{5},2) \end{gathered}$

The coordinates for the foci are: (-2-sqrt(5), 2) and (-2+sqrt(5), 2).