Let a line L _ 1 be tangent to the hyperbola x ^ 2 16 - y ^ 2 4 = 1...

The equation of tangent to the given hyperbola is $y = mx \pm \sqrt{16{m}^{2} - 4} . . . \left(i\right)</p><p>Hence,${l}_{1}: y = mx \pm \sqrt{16{m}^{2} - 4}$</p><p>Given that,${l}_{2}$is a straight line passing through origin and perpendicular to${l}_{1}$.</p><p>So,${l}_{2}: y = $\dfrac{- 1}{m}$x ⇒ m = $\dfrac{- x}{y}$ . . . $\left(ii\right)$

On solving equations \left(i\right)& \left(ii\right), we get

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

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

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

$∴ \alpha + \beta = 12$