Let $m_{1}$ and $m_{2}$ be the slopes of the tangents drawn from the point $\mathrm{P}(4,1)$ to the hyperbola $H: \frac{y^{2}}{25}-\frac{x^{2}}{16}=1$. If $\mathrm{Q}$ is the point from which the tangents drawn to $\mathrm{H}$ have slopes $\left|m_{1}\right|$ and $\left|m_{2}\right|$ and they make positive intercepts $\alpha$ and $\beta$ on the $x$-axis, then $\frac{(P Q)^{2}}{\alpha \beta}$ is equal to __________.

Equation of tangent to the hyperbola $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$ <br/><br/>$y=m x \pm \sqrt{a^2-b^2 m^2}$ <br/><br/>Given the hyperbola $H: \frac{y^2}{25} - \frac{x^2}{16} = 1$, the equation of the tangent to this hyperbola can be written as : <br/><br/>$y=mx \pm \sqrt{25 - 16m^2}$ <br/><br/>We know that the tangents pass through the point $P(4, 1)$, which gives us the equation : <br/><br/>$1 = 4m \pm \sqrt{25 - 16m^2}$ <br/><br/>Squaring both sides to get rid of the square root, we obtain : <br/><br/>$(4m - 1)^2 = 25 - 16m^2$ <br/><br/>which simplifies to the quadratic equation : <br/><br/>$4m^2 - m - 3 = 0.$ <br/><br/>Solving this equation, we find the roots $m_1 = 1$ and $m_2 = -\frac{3}{4}$, which are the slopes of the tangents. <br/><br/>Given that we are interested in the positive values of the slopes, we consider $|m_1| = 1$ and $|m_2| = \frac{3}{4}$. <br/><br/>The equations of the tangents are then : <br/><br/>1) $y = x - 3$ and <br/><br/>2) $y = \frac{3}{4}x - 4$. <br/><br/>The x-intercepts of these lines (when $y = 0$) are given by $x = \alpha = 3$ for the first line and $x = \beta = \frac{16}{3}$ for the second line. <br/><br/>The intersection point $Q$ of these tangents is found by solving the system of equations $y = x - 3$ and $y = \frac{3}{4}x - 4$, which gives $Q(-4, -7)$. <br/><br/>The square of the distance $PQ$ is then $(-4-4)^2 + (-7-1)^2 = 128$. <br/><br/>Therefore, $\frac{(PQ)^2}{\alpha \beta} = \frac{128}{3 \times \frac{16}{3}} = 8$