If $x = a (θ + \sin θ), y = a (1 - \cos θ)$ , then $y_{2}$ at $θ = \frac{π}{4}$ , is

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$\frac{2 (3 - 2 \sqrt{2})}{a}$

$\sqrt{2} - 1$

$\frac{\sqrt{2} + 1}{\sqrt{2} - 1}$

$\sqrt{2} + 1$

answer is C.

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Detailed Solution

$\frac{d y}{d x} = \frac{d y / d θ}{d x / d θ} = \frac{a \sin θ}{a (1 + \cos θ)} = \frac{2 \sin \frac{θ}{2} \cos \frac{θ}{2}}{2 \cos^{2} \frac{θ}{2}} = \tan \frac{θ}{2} \Rightarrow \frac{d y}{d x} = \tan \frac{θ}{2} \Rightarrow \frac{d^{2} y}{d x^{2}} = \frac{d}{d x} (\frac{d y}{d x}) \frac{d θ}{d x} = \frac{d}{d x} (\tan \frac{θ}{2}) \frac{1}{a (1 + \cos θ)} \Rightarrow y_{2} = (\frac{1}{2} s e c^{2} \frac{θ}{2}) \frac{1}{a (1 + \cos θ)} \Rightarrow y_{2} = \frac{1}{a {(1 + \cos θ)}^{2}} s u b s t i t u t e θ = \frac{π}{4} \Rightarrow y_{2} = \frac{1}{a {(1 + \frac{1}{\sqrt{2}})}^{2}} = \frac{2 (3 - 2 \sqrt{2})}{a}$