Let a line L_{1} be tangent to the hyperbola | vedclass

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Question

$\frac{x \sec 	heta}{4}-\frac{y 	an 	heta}{2}=1$

$m_{1}=\frac{\sec 	heta 	imes 2}{4(	an 	heta)}=\frac{\sec 	heta}{2 	an 	heta}$

$m_{2}

Vedclass · Accepted Answer

$12$

Vedclass · Answer

$\frac{x \sec 	heta}{4}-\frac{y 	an 	heta}{2}=1$

$m_{1}=\frac{\sec 	heta 	imes 2}{4(	an 	heta)}=\frac{\sec 	heta}{2 	an 	heta}$

$m_{2}=\frac{k}{h}$

$m_{1} m_{2}=-1$

$\frac{ k }{ h } \frac{\sec 	heta}{2 	an 	heta}=-1$

$\frac{ k }{2 h \sin 	heta}=-1$

$\sin 	heta=\frac{- k }{2 h } \quad \cos 	heta=\frac{\sqrt{4 h ^{2}- k ^{2}}}{2 h }$

also

$\frac{ h \sec 	heta}{4}-\frac{ k 	an 	heta}{2}=1$

$\frac{ h }{4} \frac{2 h }{\sqrt{4 h ^{2}- k ^{2}}}-\frac{ k }{2}\left(\frac{- k }{\sqrt{4 h ^{2}- k ^{2}}}ight)=1$

$h ^{2}+ k ^{2}=2 \sqrt{4 h ^{2}- k ^{2}}$

$\left( x ^{2}+ y ^{2}ight)^{2}=4\left(4 x ^{2}- y ^{2}ight)$

$\left( x ^{2}+ y ^{2}ight)^{2}=16 x ^{2}-4 y ^{2}$

$\alpha=16, \beta=-4$

$\alpha+\beta=16-4=12$

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