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

<p>Given:</p> <p><p>Transverse axis length: $2a$</p></p> <p><p>Conjugate axis length: $2b$</p></p> <p><p>One focus at $(-5, 0)$</p></p> <p><p>Directrix given by $5x + 9 = 0$</p></p> <p>The equations used are as follows:</p> <p><strong>Relationship between focus and directrix:</strong></p> <p><p>The focal length $ae = 5$</p></p> <p><p>The directrix gives $\frac{a}{e} = \frac{9}{5}$</p> <p>Solving these equations, we get:</p> <p>$ ae = 5, \quad \frac{a}{e} = \frac{9}{5} $</p></p> <p><p>From $ae = 5$, we have $a = \frac{5}{e}$.</p></p> <p><p>Substituting $a = 3$ and $e = \frac{5}{3}$.</p></p> <p><strong>Finding $ b $:</strong></p> <p><p>Use the relationship $b^2 = a^2(e^2 - 1)$:</p> <p>$ \text{Given } a = 3 \quad \text{and } e = \frac{5}{3}, \quad b = 4 $</p></p> <p><strong>Equation of the hyperbola:</strong></p> <p><p>The standard equation, after substituting values of $a$ and $b$, is:</p> <p>$ \frac{x^2}{9} - \frac{y^2}{16} = 1 $</p></p> <p><strong>Focal distances product calculation:</strong></p> <p><p>For any point $(\alpha, 2\sqrt{5})$ that lies on the hyperbola:</p> <p>$ \frac{\alpha^2}{9} - \frac{(2\sqrt{5})^2}{16} = 1 $</p></p> <p><p>Solving this gives:</p> <p>$ \alpha^2 = 9 \times \frac{36}{16} $</p></p> <p><p><strong>Product of focal distances $ \mathrm{PF}_1 \cdot \mathrm{PF}_2 $:</strong></p> <p>$ \mathrm{PF}_1 \cdot \mathrm{PF}_2 = (e\alpha - a)(e\alpha + a) = e^2\alpha^2 - a^2 $</p></p> <p><p>Substituting the values:</p> <p>$ P = e^2\alpha^2 - a^2 = \frac{25}{9} \cdot 9 \cdot \frac{9}{4} - 9 = \frac{189}{4} $</p></p> <p>Finally, to find $4p$:</p> <p>$ 4p = 4 \times \frac{189}{4} = 189 $</p>