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The normal to the rectangular hyperbola $xy=9$ at$(6,\frac{3}{2})$ meets the curve again at

\[\begin{array}{1 1}(1)\bigg(\frac{3}{8},24\bigg)&(2)\bigg(-24,\frac{-3}{8}\bigg)\\(3)\bigg(\frac{-3}{8},-24\bigg)&(4)\bigg(24,\frac{3}{8}\bigg)\end{array}\]

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The equation of the given hyperbola is $xy= 9$
$( 6,\large\frac{3}{2})$ is a point on it
$(ct_1,\large\frac{c}{t_1} )$$ =(6,\large\frac{3}{2})$
$ct_1=6,= \large\frac{c}{t_1} =\frac{3}{2}$
$c=3,t_1= 2$
Let the normal at $(ct, \large\frac{c}{t_1})$ meets the curve again at $(ct_2, \large\frac{c}{t_2})$ the $t_1^3t_2=-1$
$t_2= \large\frac{-1}{t_1^3}=\frac{-1}{8}$
The required point is $\bigg(\frac{-3}{8},-24\bigg)$
Hence 3 is the correct answer.
answered May 16, 2014 by meena.p
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