Answer
$\dfrac{(x-3)^2}{12}+\dfrac{y^2}{16}=1$
Work Step by Step
Given: Center:$(h,k)$ and $c^2=a^2-b^2$
$2a=8 \implies a=4$
and $c=2-k==2$
and $c^2=a^2-b^2 \implies (2)^2=(4)^2-b^2$
or, $b^2=16-4 =12$
Now,
$\dfrac{(x-3)^2}{12}+\dfrac{(y-0)^2}{4^2}=1$
Hence, $\dfrac{(x-3)^2}{12}+\dfrac{y^2}{16}=1$
