Question

determine the equation of the ellipse with center (2,2), focus (2,0), and vertex (2,5).multiple choices

Answer 1

Given:

The ellipse has center (2,2), focus (2,0), and vertex (2,5).

The equation of elllipse is given as,

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

a is the distance between vertex and center.

$\begin{gathered} a=\sqrt[]{(2-2)^2+(5-2)^2} \\ a=\sqrt[]{3^2}=3 \end{gathered}$

c is distance between focus and center.

$\begin{gathered} c=\sqrt[]{(2-2)^2+(0-2)^2} \\ c=\sqrt[]{4} \\ c=2 \end{gathered}$

It gives,

$\begin{gathered} c^2=a^2-b^2 \\ 2^2=a^2-b^2 \\ b^2=9-4 \\ b^2=5 \\ b=\pm\sqrt[]{5} \\ b=\sqrt[]{5}\ldots\ldots\text{ Since b is distance and it should be positive} \end{gathered}$

So, the equation of the ellipse is,

$\begin{gathered} \frac{(x-h)^2}{b^2}+\frac{(y-k)^2}{a^2}^{}=1 \\ \frac{(x-2)^2}{(\sqrt[]{5})^2^{}}+\frac{(y-2)^2}{3^2}=1 \\ \frac{(x-2)^2}{5^{}}+\frac{(y-2)^2}{9^{}}=1 \end{gathered}$

Answer: option b)