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The product of the perpendiculars drawn from the point $(8 ,0)$ on the hyperbola to its asymptotes is $\large\frac{x^{2}}{64}-\frac{y^{2}}{36}=$$1$ is

\[\begin{array}{1 1}(1)\frac{25}{576}&(2)\frac{576}{25}\\(3)\frac{6}{25}&(4)\frac{25}{6}\end{array}\]

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The equation of the hyperbola is
$\large\frac{x62}{64} -\frac{y^2}{36}$
$a^2= 64,b^2= 36$
Required length of the perpendicular $=\large\frac{a^2b^2}{a^2+b^2}$
$\qquad= \large\frac{64 \times 36}{64 +36}$
$\qquad= \large\frac{576}{25}$
Hence 2 is the correct answer.
answered May 16, 2014 by meena.p
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