Answer
$\dfrac{x^2}{25}+\dfrac{y^2}{21}=1$
See graph
Work Step by Step
We are given the ellipse:
Vertices: $(-5,0),(5,0)$
$c=2$
Because the $y$-coordinates of the center and vertex are the same, the ellipse has the equation:
$\dfrac{(x-h)^2}{a^2}+\dfrac{(y-k)^2}{b^2}=1$
Determine $h,k,a$ using the vertices:
$(h-a,k)=(-5,0)\Rightarrow h-a=-5,k=0$
$(h+a,k)=(5,0)\Rightarrow h+a=5$
$\begin{cases}
h-a=-5\\
h+a=5
\end{cases}$
$h-a+h+a=-5+5$
$2h=0$
$h=0$
$h+a=5$
$0+a=5$
$a=5$
Determine $b$:
$a^2=b^2+c^2$
$b^2=a^2-c^2$
$b^2=5^2-2^2$
$b^2=21$
$b=\sqrt{21}$
The equation of the ellipse is:
$\dfrac{x^2}{5^2}+\dfrac{y^2}{(\sqrt{21})^2}=1$
$\dfrac{x^2}{25}+\dfrac{y^2}{21}=1$
The center is:
$(h,k)=(0,0)$
Graph the ellipse:
