Question

Find the equation of the hyperbola if the centre is $(2,5)$; the distance between the directrices is $!5$; the distance between the foci is $20$ and the transverse axis is parallel to y-axis

Answer 1

Given centre of the hyperbola 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.

Distance between directrices $=\frac{2a}{e}=15$

$\Rightarrow \frac{a}{e}=\frac{15}{2}...\left(1\right)$

Distance between foci $=2ae=20$

$\Rightarrow ae=10...\left(2\right)$

From $\left(1\right)$ and $\left(2\right)$ we get

$a^2=75\Rightarrow a=5\sqrt{3}$

Therefore $e=\frac{10}{a}=\frac{10}{5\sqrt{3}}=\frac{2}{\sqrt{3}}$

$b=a\sqrt{e^2-1}=5\sqrt{3}\sqrt{\frac{4}{3}-1}=5$

We know that equation of the hyperbola is of the form $\frac{(y-k{)}^2}{a^2}-\frac{(x-h{)}^2}{b^2}=1$ where $(h,k)$ is the centre.

Thus the required equation of the hyperbola is $\frac{(y-5{)}^2}{75}-\frac{(x-2{)}^2}{25}=1$.