Answer
$\dfrac{(x-1)^2}{10}+\dfrac{(y-2)^2}{1}=1$
See graph
Work Step by Step
We are given the ellipse:
Center: $(1,2)$
Focus: $(4,2)$
Point on the graph: $(1,3)$
Because the $y$-coordinates of the 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:
$(h,k)=(1,2)$
$h=1$
$k=2$
Determine $c$ using the focus:
$(h+c,k)=(3,0)$
$(0+c,0)=(3,0)$
$(c,0)=(3,0)$
$c=3$
We get the relation between $a$ and $b$:
$a^2=b^2+c^2$
$a^2-b^2=c^2$
$a^2-b^2=3^2$
$a^2-b^2=9$
Use the point $(1,3)$ to determine another relation between $a$ and $b$:
$\dfrac{(1-1)^2}{a^2}+\dfrac{(3-2)^2}{b^2}=1$
$\dfrac{1}{b^2}=1$
$b^2=1$
$b=1$
Determine $a$:
$a^2-b^2=9$
$a^2-1^2=9$
$a^2=10$
$a=\sqrt{10}$
The equation of the ellipse is:
$\dfrac{(x-1)^2}{(\sqrt{10})^2}+\dfrac{(y-2)^2}{1^2}=1$
$\dfrac{(x-1)^2}{10}+\dfrac{(y-2)^2}{1}=1$
The center is:
$(h,k)=(1,2)$
Determine the vertices and co-vertices:
$(h-a,k)=(1-\sqrt{10},2)$
$(h+a,k)=(1+\sqrt{10},2)$
$(h,k-b)=(1,2-1)=(1,1)$
$(h,k+b)=(1,2+1)=(1,3)$
Graph the ellipse:
