The sub-tangent at any point of the curve $x^{m} y^{n} = a^{m + 1}$ varies as

$(A b s c i s s a)^{2}$

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

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Abscissa

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Ordinate

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Solution

The correct option is C
Abscissa

We have, $x^{m} y^{n} = a^{m + 1} \Rightarrow m l o g x + n l o g y = (m + 1) l o g a ∴ \frac{m}{x} + \frac{n}{y} \frac{d y}{d x} = 0 \Rightarrow \frac{d x}{d y} = \frac{n x}{m y} ∴ S u b - t a n g e n t == ∣ ∣ y \frac{d x}{d y} ∣ ∣ = \frac{n x}{m} a x .$
Hence (c) is the correct answer.

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