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(B) Let $(a \sec 	heta, b 	an 	heta)$ be any point on the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$. The equations of the asymptotes are $\frac

Vedclass · Accepted Answer

$\frac{a^2 b^2}{a^2+b^2}$

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(B) Let $(a \sec 	heta, b 	an 	heta)$ be any point on the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$. The equations of the asymptotes are $\frac{x}{a} + \frac{y}{b} = 0$ and $\frac{x}{a} - \frac{y}{b} = 0$. The perpendicular distance $P_1$ from $(a \sec 	heta, b 	an 	heta)$ to the line $\frac{x}{a} + \frac{y}{b} = 0$ is given by: $P_1 = \frac{|\sec 	heta + 	an 	heta|}{\sqrt{\frac{1}{a^2} + \frac{1}{b^2}}}$. The perpendicular distance $P_2$ from $(a \sec 	heta, b 	an 	heta)$ to the line $\frac{x}{a} - \frac{y}{b} = 0$ is given by: $P_2 = \frac{|\sec 	heta - 	an 	heta|}{\sqrt{\frac{1}{a^2} + \frac{1}{b^2}}}$. The product $P_1 P_2$ is: $P_1 P_2 = \frac{|\sec^2 	heta - 	an^2 	heta|}{\frac{1}{a^2} + \frac{1}{b^2}} = \frac{1}{\frac{a^2 + b^2}{a^2 b^2}} = \frac{a^2 b^2}{a^2 + b^2}$.

The product of the perpendicular distances from any point on the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ to its asymptotes is

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