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
$x=2\cos t+2t\sin t\Rightarrow y=2\sin t-2t\cos t,\frac{dx}{dy}$
$=\frac{-2\sin t+2t\cos t+2\sin t}{2\cos t+2t\sin t-2\cos t}=\cot t$
$\Rightarrow {\left(\frac{-dx}{dy}\right)}_{k=\pi /4}=-1$
$\left(x_1,y_l\right)\equiv \left(\sqrt{2}+2\left(\frac{\pi }{4}\right)\right)\frac{1}{\sqrt{2}},\sqrt{2}-2\left(\frac{\pi }{4}\right)\frac{1}{\sqrt{2}}$
$\equiv \left(\sqrt{2}+\frac{\pi }{2\sqrt{2}},\sqrt{2}-\frac{\pi }{2\sqrt{2}}\right)$
$\therefore$ 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
$=\frac{|0+0-2\sqrt{2}|}{\sqrt{1+1}}=2$