Let a line $L_1$ be tangent to the hyperbola $\frac{x^2}{16}-\frac{y^2}{4}=1$ and let $L_2$ be the line passing through the origin and perpendicular to $L_1$. If the locus of the point of intersection of $L_1$ and $L_2$ is ${\left(x^2+y^2\right)}^2=$ $\alpha x^2+\beta y^2$, then $\alpha +\beta$ is equal to ______.

The equation of tangent to the given hyperbola is $y=mx\pm \sqrt{16m^2-4} ...\left(\mathrm{i}\right)$

Hence, $l_1:y=mx\pm \sqrt{16m^2-4}$

Given that, $l_2$ is a straight line passing through origin and perpendicular to $l_1$.

So, $l_2:y=\frac{-1}{m}x$  $\Rightarrow m=\frac{-x}{y} ...\left(\mathrm{ii}\right)$

On solving equations $\left(\mathrm{i}\right)$ & $\left(\mathrm{ii}\right)$, we get

$y=\left(\frac{-x}{y}\right)x\pm \sqrt{16{\left(\frac{-x}{y}\right)}^2-4}$  $\Rightarrow y=\frac{-x^2}{y}\pm \frac{\sqrt{16x^2-4y^2}}{y}$

$\Rightarrow {\left(y^2+x^2\right)}^2=16x^2-4y^2$

On comparing the above equation with ${\left(x^2+y^2\right)}^2=\alpha x^2+\beta y^2$, we get $\alpha =16, \beta =-4$   

$\therefore \alpha +\beta =12$