The length of the normal at point ‘t’ of the curve x = a(t + sin t), y = a (1 - cos t) is
[RPET 2001]

a sin t

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2 a s i n^{3} (t / 2) s e c (t / 2)

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2a sin(t / 2) tan(t/2)

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2a sin (t / 2)

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Solution

The correct option is B $2 a s i n^{3} (t / 2) s e c (t / 2)$
$y = a x^{n + 1} + b x^{- n} \Rightarrow \frac{d y}{d x} = (n + 1) a x^{n} - n b x^{- n - 1}$
$\Rightarrow \frac{d^{2} y}{d x^{2}} = n (n + 1) a x^{n - 1} - n (n + 1) b x^{- n - 2}$
$\Rightarrow x^{2} \frac{d^{2} y}{d x^{2}} = n (n + 1) y .$

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