Let the lengths of the transverse and conjugate axes of a hyperbola in standard form be 2a and 2b, respectively, and one focus and the corresponding directrix of this hyperbola be $$(-5, 0)$$ and $$5x + 9 = 0$$, respectively. If the product of the focal distances of a point $$\left(\alpha, 2\sqrt{5}\right)$$ on the hyperbola is p, then 4p is equal to _____.

The standard horizontal hyperbola with centre at the origin is written as $$\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$$.
Its basic facts are:

• Foci $$\left(\pm c,\,0\right)$$ where $$c^{2}=a^{2}+b^{2}$$.
• Eccentricity $$e=\frac{c}{a}\ (e\gt 1)$$.
• Directrices $$x=\pm\frac{a}{e}$$ (each focus has the nearer directrix).

We are told that one focus is $$(-5,0)$$ and the corresponding directrix is $$5x+9=0$$, i.e. $$x=-\frac{9}{5}$$.
Hence

$$c=5,\qquad -\frac{a}{e}=-\frac{9}{5}\; \Longrightarrow\; \frac{a}{e}=\frac{9}{5}\; -(1)$$

Using $$e=\frac{c}{a}$$, substitute $$c=5$$ into $$e=\frac{5}{a}$$ and then into $$(1)$$:

$$\frac{a}{e}=a\left(\frac{a}{5}\right)=\frac{a^{2}}{5}=\frac{9}{5}\; \Longrightarrow\; a^{2}=9\; \Longrightarrow\; a=3.$$

Now compute $$b^{2}$$ using $$c^{2}=a^{2}+b^{2}$$:

$$25=9+b^{2}\; \Longrightarrow\; b^{2}=16\; \Longrightarrow\; b=4.$$

Therefore the hyperbola is $$\frac{x^{2}}{9}-\frac{y^{2}}{16}=1$$.

The given point $$\left(\alpha,\,2\sqrt{5}\right)$$ lies on the curve, so substitute it:

$$\frac{\alpha^{2}}{9}-\frac{(2\sqrt{5})^{2}}{16}=1 \quad\Longrightarrow\quad \frac{\alpha^{2}}{9}-\frac{20}{16}=1.$$

Simplify:

$$\frac{\alpha^{2}}{9}-\frac{5}{4}=1\; \Longrightarrow\; \frac{\alpha^{2}}{9}=\frac{9}{4}\; \Longrightarrow\; \alpha^{2}=\frac{81}{4}\; \Longrightarrow\; \alpha=\pm\frac{9}{2}.$$

Let $$P(\alpha,2\sqrt{5})$$ be the point; its distances to the two foci $$F_{1}(5,0)$$ and $$F_{2}(-5,0)$$ are

$$PF_{1}=\sqrt{(\alpha-5)^{2}+(2\sqrt{5})^{2}},\qquad PF_{2}=\sqrt{(\alpha+5)^{2}+(2\sqrt{5})^{2}}.$$

The required product is $$p=PF_{1}\,PF_{2}=\sqrt{\bigl[(\alpha-5)^{2}+20\bigr]\bigl[(\alpha+5)^{2}+20\bigr]}.$$

$$\alpha=\frac{9}{2}=4.5$$

$$\bigl[(\alpha-5)^{2}+20\bigr]=(4.5-5)^{2}+20=0.25+20=20.25,$$ $$\bigl[(\alpha+5)^{2}+20\bigr]=(4.5+5)^{2}+20=90.25+20=110.25.$$

Product inside the root: $$20.25\times110.25=2232.5625,$$ hence

$$p=\sqrt{2232.5625}=47.25.$$

$$\alpha=-\frac{9}{2}=-4.5$$

This merely interchanges the two factors, giving the same product $$p=47.25$$.

Finally, $$4p=4\times47.25=189.$$

Hence, $$\boxed{4p=189}$$.