Question

The distance, from the origin, of the normal to the curve, $x = 2 \,\cos t + 2t \,\sin t$,
$y = 2 \,\sin t - 2t$ at $t = \frac{\pi}{4}, $ is :

• A. $4$
• B. $2\sqrt{2}$
• C. $2$
• D. $\sqrt{2}$

Answer 1

Answer: (C)

Parametric form of curve is
$x=2 \cos t+2 t \sin t \Rightarrow y=2 \sin t-2 t \cos t, \frac{d x}{d y}$
$=\frac{-2 \sin t+2 t \cos t+2 \sin t}{2 \cos t+2 t \sin t-2 \cos t}=\cot t$
$\Rightarrow \left(\frac{-d x}{d y}\right)_{k=\pi / 4}=-1$
$\left(x_{1}, y_{ l }\right) \equiv\left(\sqrt{2}+2\left(\frac{\pi}{4}\right)\right) \frac{1}{\sqrt{2}}, \sqrt{2}-2\left(\frac{\pi}{4}\right) \frac{1}{\sqrt{2}}$
$\equiv\left(\sqrt{2}+\frac{\pi}{2 \sqrt{2}}, \sqrt{2}-\frac{\pi}{2 \sqrt{2}}\right)$
$\therefore $ Equation of normal at $t=\pi / 4$ will be
$\left[y-\left(\sqrt{2}-\frac{\pi}{2 \sqrt{2}}\right)\right]=(-1)\left[x-\left(\sqrt{2}+\frac{\pi}{2 \sqrt{2}}\right)\right]$
or $x+y-2 \sqrt{2}=0 \Rightarrow $ Its distance from origin is
$=\frac{|0+0-2 \sqrt{2}|}{\sqrt{1+1}}=2$