Answer
$\dfrac{(x-2)^2}{25}+\dfrac{(y+2)^2}{21}=1$
See graph
Work Step by Step
We are given the ellipse:
Center: $(2,-2)$
Focus: $(4,-2)$
Vertex: $(7,-2)$
Because the $y$-coordinates of the vertex, center, and focus are the same, the ellipse has the equation:
$\dfrac{(x-h)^2}{a^2}+\dfrac{(y-k)^2}{b^2}=1$
Determine $h,k$ using the center:
$(h,k)=(2,-2)$
$h=2$
$k=-2$
Determine $a$ using the vertex:
$(h+a,k)=(7,-2)$
$(2+a,-2)=(7,-2)$
$2+a=7$
$a=5$
Determine $c$ using the focus:
$(h+c,k)=(4,-2)$
$(2+c,-2)=(4,-2)$
$2+c=4$
$c=2$
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:
$\dfrac{(x-2)^2}{5^2}+\dfrac{(y+2)^2}{(\sqrt{21})^2}=1$
$\dfrac{(x-2)^2}{25}+\dfrac{(y+2)^2}{21}=1$
The center is:
$(h,k)=(2,-2)$
Determine the vertices and co-vertices:
$(h-a,k)=(2-5,-2)=(-3,-2)$
$(h+a,k)=(7,-2)$
$h,k-b)=(2,-2-\sqrt{21})$
$h,k+b)=(2,-2+\sqrt{21})$
Graph the ellipse:
