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

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

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

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

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

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Solution

The correct option is A
$(A b c i s s a)^{- 3}$

We have, $x^{2} y^{2} = a^{2} (x^{2} - a^{2}) . . . (1) \Rightarrow x^{2} . 2 y \frac{d y}{d x} + y^{2} . 2 x - a^{2} . 2 x \Rightarrow \frac{d y}{d x} = \frac{a^{2} - y^{2}}{x y} . ∴ S u b - n o r m a l = y \frac{d y}{d x} = \frac{a^{2} - y^{2}}{x y} = \frac{x^{2} (a^{2} - y^{2})}{x^{3}} = \frac{a^{4}}{x^{3}} [∵ f r o m (1) x^{2} (a^{2} - y^{2}) = a^{4}]$
The sub-normal varies inversely as the cube of its abcissa.
Hence (a) is the correct answer.

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