The sub-normal at any point of the curve $x^{2} y^{2} = a^{2} (x^{2} - a^{2})$ varies as

{(a b s c i s s a)}^{- 3}

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{(a b s c i s s a)}^{3}

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{(o r d i n a t e)}^{- 3}

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None of these

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Solution

The correct option is A ${(a b s c i s s a)}^{- 3}$
We have, $x^{2} y^{2} = a^{2} (x^{2} - a^{2})$ ...(1)
$\Rightarrow x^{2} .2 y \frac{d y}{d x} + y^{2} .2 x = a^{2} .2 x$
$\Rightarrow \frac{d y}{d x} = \frac{a^{2} - y^{2}}{x y} .$
$∴$ Sub-normal $= y . m = y \frac{d y}{d x} = \frac{a^{2} - y^{2}}{x} = \frac{x^{2} (a^{2} - y^{2})}{x^{3}}$
$= \frac{a^{4}}{x^{3}}$ $[∵$ from (1) $x^{2} (a^{2} - y^{2}) = a^{4}]$
$\Rightarrow$ The sub-normal varies inversely as the cube of its ab-scissa.

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