The product of the lengths of perpendiculars | vedclass

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Question

Given          $\frac{x^{2}}{2}-\frac{y^{2}}{1}=1$

Here,         $a^{2}=2, \quad b^{2}=1$

Equa

Vedclass · Accepted Answer

$\frac{2}{3}$

Vedclass · Answer

Given          $\frac{x^{2}}{2}-\frac{y^{2}}{1}=1$

Here,         $a^{2}=2, \quad b^{2}=1$

Equation of asymptotes to the given hyperbola is

$\frac{x}{{\sqrt 2 }} - \frac{y}{1} = 0$ and ${m{ }}\frac{x}{{\sqrt 2 }} + \frac{y}{1} = 0$

Let $P(\sqrt{2} \sec 	heta, 	an 	heta)$ be any point, then product of length of perpendicular

${=\frac{\left[\frac{\sqrt{2} \sec 	heta}{\sqrt{2}}-\frac{	an 	heta}{1}ight]\left[\frac{\sqrt{2} \sec 	heta}{\sqrt{2}}+\frac{	an 	heta}{1}ight]}{\sqrt{\frac{1}{2}+\frac{1}{1}} \sqrt{\frac{1}{2}+\frac{1}{1}}}} $

${=\frac{\sec ^{2} 	heta-	an ^{2} 	heta}{\frac{3}{2}}=\frac{2}{3}}$

The product of the lengths of perpendiculars drawn from any point on the hyperbola $x^2 -2y^2 -2=0$ to its asymptotes is

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