The length of sub-tangent to the curve $y^{n} = a^{n - 1} . x$ at (x,y) on it is

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$|nx|$

$n^{2} |x|$

$\frac{n^{2}}{|x|}$

$|\frac{n}{x}|$

answer is B.

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Detailed Solution

Given $y^{n} = a^{x - 1} x$
$m = {(\frac{dy}{dx})}_{p (x, y)} = \frac{a^{n - 1}}{n y^{n - 1}}$
L.S.T: $|\frac{y}{m}| = |y \frac{n y^{n - 1}}{a^{n - 1}}| = |\frac{n y^{n}}{a^{n - 1}}| \Rightarrow |\frac{n a^{n - 1} x}{a^{n - 1}}| \Rightarrow |n x|$

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