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

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(B) The equation of the tangent $L_{1}$ to the hyperbola $\frac{x^{2}}{16}-\frac{y^{2}}{4}=1$ at point $(4 \sec 	heta, 2 	an 	heta)$ is $\frac{x \s

Vedclass · Accepted Answer

$12$

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(B) The equation of the tangent $L_{1}$ to the hyperbola $\frac{x^{2}}{16}-\frac{y^{2}}{4}=1$ at point $(4 \sec 	heta, 2 	an 	heta)$ is $\frac{x \sec 	heta}{4}-\frac{y 	an 	heta}{2}=1$.The slope of $L_{1}$ is $m_{1} = \frac{\sec 	heta}{2 	an 	heta}$.Let the point of intersection be $(h, k)$. Since $L_{2}$ passes through $(0, 0)$ and $(h, k)$,its slope is $m_{2} = \frac{k}{h}$.Since $L_{1} \perp L_{2}$,we have $m_{1} m_{2} = -1$,so $\frac{k}{h} \cdot \frac{\sec 	heta}{2 	an 	heta} = -1$,which gives $\frac{k}{2h \sin 	heta} = -1$,or $\sin 	heta = -\frac{k}{2h}$.Then $\cos^{2} 	heta = 1 - \sin^{2} 	heta = 1 - \frac{k^{2}}{4h^{2}} = \frac{4h^{2}-k^{2}}{4h^{2}}$,so $\cos 	heta = \frac{\sqrt{4h^{2}-k^{2}}}{2h}$.Substituting these into the tangent equation $\frac{h \sec 	heta}{4} - \frac{k 	an 	heta}{2} = 1$:$\frac{h}{4} \cdot \frac{2h}{\sqrt{4h^{2}-k^{2}}} - \frac{k}{2} \cdot \left( \frac{-k}{\sqrt{4h^{2}-k^{2}}} ight) = 1$$\frac{2h^{2}+k^{2}}{2\sqrt{4h^{2}-k^{2}}} = 1 \implies 2h^{2}+k^{2} = 2\sqrt{4h^{2}-k^{2}}$Squaring both sides: $(2h^{2}+k^{2})^{2} = 4(4h^{2}-k^{2}) = 16h^{2}-4k^{2}$.Replacing $(h, k)$ with $(x, y)$,we get $(2x^{2}+y^{2})^{2} = 16x^{2}-4y^{2}$.Wait,the standard locus of the foot of the perpendicular from the origin to a tangent of a hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ is $(x^{2}+y^{2})^{2} = a^{2}x^{2}-b^{2}y^{2}$.Here $a^{2}=16$ and $b^{2}=4$,so the locus is $(x^{2}+y^{2})^{2} = 16x^{2}-4y^{2}$.Thus $\alpha=16$ and $\beta=-4$.$\alpha+\beta = 16-4 = 12$.

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