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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
$11$
$12$
$15$
$16$
Solution
$\frac{x \sec \theta}{4}-\frac{y \tan \theta}{2}=1$
$m_{1}=\frac{\sec \theta \times 2}{4(\tan \theta)}=\frac{\sec \theta}{2 \tan \theta}$
$m_{2}=\frac{k}{h}$
$m_{1} m_{2}=-1$
$\frac{ k }{ h } \frac{\sec \theta}{2 \tan \theta}=-1$
$\frac{ k }{2 h \sin \theta}=-1$
$\sin \theta=\frac{- k }{2 h } \quad \cos \theta=\frac{\sqrt{4 h ^{2}- k ^{2}}}{2 h }$
also
$\frac{ h \sec \theta}{4}-\frac{ k \tan \theta}{2}=1$
$\frac{ h }{4} \frac{2 h }{\sqrt{4 h ^{2}- k ^{2}}}-\frac{ k }{2}\left(\frac{- k }{\sqrt{4 h ^{2}- k ^{2}}}\right)=1$
$h ^{2}+ k ^{2}=2 \sqrt{4 h ^{2}- k ^{2}}$
$\left( x ^{2}+ y ^{2}\right)^{2}=4\left(4 x ^{2}- y ^{2}\right)$
$\left( x ^{2}+ y ^{2}\right)^{2}=16 x ^{2}-4 y ^{2}$
$\alpha=16, \beta=-4$
$\alpha+\beta=16-4=12$
