1. Write the hyperbola in standard form

Since the given focus is $(-5,0)$ and the corresponding directrix is of the form $5x+9=0$, the hyperbola is centered at the origin with transverse axis along the $x$-axis.

So its standard equation is

with foci at $(\pm c,0)$, where

Given one focus is $(-5,0)$, hence $c=5.$

2. Use the directrix equation

For the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1,$ the directrices are $x=\pm \frac{a}{e}=\pm \frac{a^2}{c}.$

The given directrix is $5x+9=0 \implies x=-\frac{9}{5}.$ Since this corresponds to the focus $(-5,0)$, we take $-\frac{a^2}{c}=-\frac{9}{5}.$ With $c=5$, $\frac{a^2}{5}=\frac{9}{5} \implies a^2=9 \implies a=3.$

Then $b^2=c^2-a^2=25-9=16.$

So the hyperbola is

3. Find the point on the hyperbola

The point $(\alpha,2\sqrt5)$ lies on the hyperbola. Substitute $y=2\sqrt5$:

Since $(2\sqrt5)^2=20$,

So $\alpha=\pm \frac{9}{2}.$

4. Use the product of focal distances property

For any point $(x,y)$ on the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1,$ the product of its distances from the two foci is

but the standard simpler identity is

A more direct route is to compute using coordinates.

Let the foci be $F_1(-5,0)$ and $F_2(5,0)$, and point be $P(\alpha,2\sqrt5)$.

Then $PF_1^2=(\alpha+5)^2+20,$ $PF_2^2=(\alpha-5)^2+20.$

Therefore $p^2=\left[(\alpha+5)^2+20\right]\left[(\alpha-5)^2+20\right].$

Expand:

Using $\alpha^2=\frac{81}{4}$,

Hence

So

5. Find $4p$

Final Answer