Let $m_1$ and $m_2$ be the slopes of the tangents drawn from the point $P\left(4,1\right)$ to the hyperbola $H:\frac{y^2}{25}-\frac{x^2}{16}=1$ If $Q$ is the point from which the tangents drawn to $H$ have slopes $\left|m_1\right|$ and $\left|m_2\right|$ and they make positive intercepts $\alpha$ and $\beta$ on the $x-$axis, then $\frac{(PQ{)}^2}{\alpha \beta }$ is equal to $\_\_\_\_\_\_\_.$

Given that $H:\frac{y^2}{25}-\frac{x^2}{16}=1$

Equation of tangent to the hyperbola is$:y=mx+\sqrt{25-16m^2}$

Passes through $\left(4,1\right)$

$\Rightarrow y=mx+\sqrt{25-16m^2}$

$\Rightarrow 1=4m+\sqrt{25-16m^2}$

$\Rightarrow 4m^2-m-3$

$\Rightarrow m_1=1,m_2=\frac{-3}{4}$

$\Rightarrow \left|m_1\right|=1, \left|m_2\right|=\frac{3}{4}$

Equation of tangents with slopes $1$ & $\frac{3}{4}$

$\Rightarrow y=x-3.......\left(\mathrm{i}\right)$

$\Rightarrow y=\frac{3}{4}x-4......\left(\mathrm{ii}\right)$

$\left(\mathrm{i}\right)$ & $\left(\mathrm{ii}\right)$ $\Rightarrow Q\equiv (-4,-7)$

$P\equiv (4,1)$

$\Rightarrow (PQ{)}^2=8^2+8^2=128$

On converting $\left(i\right)$ and $\left(ii\right)$ in intercept form we get,

$\left(\mathrm{i}\right)\Rightarrow \alpha =3, \left(\mathrm{ii}\right)\Rightarrow \beta =\frac{16}{3}$

$\Rightarrow \frac{(PQ{)}^2}{\alpha \beta }=\frac{128}{3\times \frac{16}{3}}=8$

Hence this is the required option.