Question

Write the equation of the ellipse that has a center at (5,6), a focus at (-1,6), and a vertex at (-2,6).

Answer 1

Answer:

$\dfrac{(x-5)^2}{49}+\dfrac{(y-6)^2}{13}=1$

Step-by-step explanation:

The equation of the ellipse can be found in the form

$\dfrac{(x-x_0)^2}{a^2}+\dfrac{(y-y_0)^2}{b^2}=1,$

where $(x_0,y_0)$ is the center of the ellipse.

The distance from the center to the focus is c:

$c=\sqrt{(5-(-1))^2+(6-6)^2}=\sqrt{36}=6.$

The distance from the center to the vertex is a:

$a=\sqrt{(5-(-2))^2+(6-6)^2}=\sqrt{49}=7.$

Since

$c^2=a^2-b^2,$

we have that

$36=49-b^2,\\ \\b^2=49-36=13.$

Hence, the equation of the ellipse is

$\dfrac{(x-5)^2}{49}+\dfrac{(y-6)^2}{13}=1.$