Question

This ellipse is centered at theorigin. Find its equation.Vertices: (-4,0) and (4,0)Foci: (-3,0) and (3,0)x² y²+[?]1

Answer 1

Explanation

We are given the following:

`\begin{gathered} Center(h,k)\to(0,0) \\ Vertices:(-4,0)and(4,0) \\ Foci:(-3,0)and(3,0) \end{gathered}`

We are required to determine the equation of the ellipse.

We know that the general equation of an ellipse is of the form:

`((x-h)^2)/(a^2)+((y-k)^2)/(b^2)=1`

We can determine the equation of the ellipse as follows:

`\begin{gathered} Recall\text{ }that: \\ (h-4)^2=a^2 \\ (0-4)^2=a^2 \\ a^2=16 \\ \\ Also, \\ (h-3)^2=a^2-b^2 \\ (0-3)^2=16-b^2 \\ 9=16-b^2 \\ \therefore b^2=16-9=7 \\ \\ the\text{ }equation\text{ }becomes \\ ((x-h)^(2))/(a^(2))+((y-k)^(2))/(b^(2))=1 \\ ((x-0)^2)/(16)+((y-0)^2)/(7)=1 \\ (x^2)/(16)+(y^2)/(7)=1 \end{gathered}`

Hence, the answer is:

`(x^(2))/(16)+(y^(2))/(7)=1`