If 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 6 and eccentricity of the hyperbola is $\sqrt{3}$, then the length of the conjugate axis of the hyperbola is

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Aicharya · Accepted Answer

$P Q=\left|\frac{a b \sec 	heta-a b 	an 	heta}{\sqrt{b^2+a^2}}ight|$ and $P R=\left|\frac{a b \sec 	heta+a b 	an 	heta}{\sqrt{b^2+a^2}}ight|$ According to question, $\frac{a b(\sec 	heta-	an 	heta)}{\sqrt{a^2+b^2}} 	imes \frac{a b(\sec 	heta+	an 	heta)}{\sqrt{a^2+b^2}}=6$ $\Rightarrow \frac{a^2 b^2\left(\sec ^2 	heta-	an ^2 	hetaight)}{\left(a^2+b^2ight)}=6$ $\Rightarrow \quad \frac{a^2 b^2}{a^2 e^2}=6 \quad\left[\because a^2+b^2=a^2 e^2ight]$ $\begin{array}{ll}\Rightarrow & b^2=6 	imes e^2=6 	imes 3=18 \ \because & a^2+b^2=a^2 e^2\end{array}$ $	herefore \quad a^2+18=a^2 	imes 3 \Rightarrow a^2=9$ $\Rightarrow \quad a=3$ Length of transverse axis $=2 	imes 3=6$

If the product of the perpendicular distances from any point on the hyperbola…

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