Answer
$\dfrac{(x-1)^2}{1}+\dfrac{(y-2)^2}{5}=1$
See graph
Work Step by Step
We are given the ellipse:
Center: $(1,2)$
Focus: $(1,4)$
Point on the graph: $(2,2)$
Because the $x$-coordinates of the center and focus are the same, the ellipse has the equation:
$\dfrac{(x-h)^2}{b^2}+\dfrac{(y-k)^2}{a^2}=1$
Determine $h,k$ using the center:
$(h,k)=(1,2)$
$h=1$
$k=2$
Determine $c$ using the focus:
$(h,k+c)=(1,4)$
$(1,2+c)=(1,4)$
$2+c=4$
$c=2$
We get the relation between $a$ and $b$:
$a^2=b^2+c^2$
$a^2-b^2=c^2$
$a^2-b^2=2^2$
$a^2-b^2=4$
Use the point $(2,2)$ to determine another relation between $a$ and $b$:
$\dfrac{(2-1)^2}{b^2}+\dfrac{(2-2)^2}{a^2}=1$
$\dfrac{1}{b^2}=1$
$b^2=1$
$b=1$
Determine $a$:
$a^2-b^2=4$
$a^2-1^2=4$
$a^2=5$
$a=\sqrt{5}$
The equation of the ellipse is:
$\dfrac{(x-1)^2}{1^2}+\dfrac{(y-2)^2}{(\sqrt 5)^2}=1$
$\dfrac{(x-1)^2}{1}+\dfrac{(y-2)^2}{5}=1$
The center is:
$(h,k)=(1,2)$
Determine the vertices and co-vertices:
$(h,k-a)=(1,2-\sqrt{5})$
$(h,k+a)=(1,2+\sqrt{5})$
$(h-b,k)=(1-1,2)=(0,2)$
$(h+b,k)=(1+1,2)=(2,2)$
Graph the ellipse:
