Equation of the vertical hyperbolas

a) Firstly identify in the expression $\dfrac{(y-y_0)^2}{a^2}-\dfrac{(x-x_0)^2}{b^2}=1$ the equation. To do so, divide the equation given by $2$ so that the same remains in the part of the right. $$\dfrac{y^2}{4}-\dfrac{(x-4)^2}{36}=1$$ Next, identify: $$\dfrac{y^2}{2^2}-\dfrac{(x-4)^2}{6^2}=1$$ The center is in $C(x_0,y_0)$, therefore $C(4,0)$.

b) The apexes are in $F'(x_0,y_0-a)$ and $F(x_0,y_0+a)$. Like $a=2$, $F'(4,-2)$ and $F(4,2)$.

c) Look for the focal distance: $c^2=a^2+b^2=2^2+6^2=4+36=40$. Do the root: $$c=\sqrt{40}=2\sqrt{10}$$