Let a line L₁ be tangent to the hyperbola ${{{x^2}} \over {16}} - {{{y^2}} \over 4} = 1$ and let L₂ be the line passing through the origin and perpendicular to L₁. If the locus of the point of intersection of L₁ and L₂ is ${({x^2} + {y^2})^2} = \alpha {x^2} + \beta {y^2}$, then $\alpha$ + $\beta$ is equal to _____________.

<p>Equation of L₁ is</p> <p>${{x\sec \theta } \over 4} - {{y\tan \theta } \over 2} = 1$ ..... (i)</p> <p>Equation of line L₂ is</p> <p>${{x\tan \theta } \over 2} + {{y\sec \theta } \over 4} = 0$ ..... (ii)</p> <p>$\because$ Required point of intersection of L₁ and L₂ is (x₁, y₁) then</p> <p>${{{x_1}\sec \theta } \over 4} - {{{y_1}\tan \theta } \over 2} - 1 = 0$ ...... (iii)</p> <p>and ${{{y_1}\sec \theta } \over 4} - {{{x_1}\tan \theta } \over 2} = 0$ ...... (iv)</p> <p>From equations (iii) and (iv)</p> <p>$\sec \theta = {{4{x_1}} \over {x_1^2 + y_1^2}}$ and $\tan \theta = {{ - 2{y_1}} \over {x_1^2 + y_1^2}}$</p> <p>$\therefore$ Required locus of (x₁, y₁) is</p> <p>${({x^2} + {y^2})^2} = 16{x^2} - 4{y^2}$</p> <p>$\therefore$ $\alpha$ = 16, $\beta$ = $-$4</p> <p>$\therefore$ $\alpha$ + $\beta$ = 12</p>