Given:

Transverse axis length: $2a$

Conjugate axis length: $2b$

One focus at $(-5, 0)$

Directrix given by $5x + 9 = 0$

The equations used are as follows:

Relationship between focus and directrix:

The focal length $ae = 5$

The directrix gives $\frac{a}{e} = \frac{9}{5}$

Solving these equations, we get:

$ ae = 5, \quad \frac{a}{e} = \frac{9}{5} $

From $ae = 5$, we have $a = \frac{5}{e}$.

Substituting $a = 3$ and $e = \frac{5}{3}$.

Finding $ b $:

Use the relationship $b^2 = a^2(e^2 - 1)$:

$ \text{Given } a = 3 \quad \text{and } e = \frac{5}{3}, \quad b = 4 $

Equation of the hyperbola:

The standard equation, after substituting values of $a$ and $b$, is:

$ \frac{x^2}{9} - \frac{y^2}{16} = 1 $

Focal distances product calculation:

For any point $(\alpha, 2\sqrt{5})$ that lies on the hyperbola:

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

Solving this gives:

$ \alpha^2 = 9 \times \frac{36}{16} $

Product of focal distances $ \mathrm{PF}_1 \cdot \mathrm{PF}_2 $:

$ \mathrm{PF}_1 \cdot \mathrm{PF}_2 = (e\alpha - a)(e\alpha + a) = e^2\alpha^2 - a^2 $

Substituting the values:

$ P = e^2\alpha^2 - a^2 = \frac{25}{9} \cdot 9 \cdot \frac{9}{4} - 9 = \frac{189}{4} $

Finally, to find $4p$:

$ 4p = 4 \times \frac{189}{4} = 189 $