Q.Find the equation of the hyperbola if the centre is $(2, 5)$ ; the distance between the directrices is 15 ; the distance between the foci is 20 and the transverse axis is parallel to $y$-axis. OR Find the volume of the solid obtained by revolving the loop of the curve $2ay^2 = x(x-a)^2$ about $x$-axis. Here $a > 0$.

Main part: use the standard relations for a hyperbola's foci-distance and directrix-distance to find $a^2,b^2,e$, then write the equation about centre $(2,5)$. OR alternative: integrate $\pi y^2\,dx$ over the loop of the curve to get the volume of revolution.

Main part — Hyperbola

1. For a hyperbola with semi-transverse axis $a$ and eccentricity $e$: distance between foci $=2ae$, distance between directrices $=\dfrac{2a}{e}$.
2. Given $2ae=20 \Rightarrow ae=10$, and $\dfrac{2a}{e}=15 \Rightarrow \dfrac{a}{e}=7.5$.
3. Multiplying the two: $(ae)\left(\dfrac{a}{e}\right)=a^2 = 10\times7.5=75$.
4. Dividing the two: $\dfrac{ae}{a/e}=e^2=\dfrac{10}{7.5}=\dfrac43$.
5. $b^2 = a^2(e^2-1) = 75\left(\dfrac43-1\right)=75\times\dfrac13=25$.
6. Since the transverse axis is parallel to the $y$-axis and the centre is $(2,5)$, the standard form is $\dfrac{(y-5)^2}{a^2}-\dfrac{(x-2)^2}{b^2}=1$
7. Substituting: $\dfrac{(y-5)^2}{75}-\dfrac{(x-2)^2}{25}=1$.

OR — alternative: Volume of the loop of $2ay^2=x(x-a)^2$ revolved about the $x$-axis ($a>0$)

8. From the curve, $y^2 = \dfrac{x(x-a)^2}{2a}$. Since $y^2\ge0$ requires $x\ge0$, and $y=0$ at $x=0$ and at $x=a$, the loop lies between $x=0$ and $x=a$. …