Question

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

Answer 1

$x=2\cos t+2t\sin t\frac{dx}{dt}=2t\cos ty=2\sin t-2t\cos t\frac{dy}{dt}=2t\sin t\Rightarrow \frac{dy}{dx}=\tan t$
So, slope of normal at $t=\frac{\pi }{4}$ is $-\frac{1}{\tan t}=-1$
At $t=\frac{\pi }{4},$ $x=\sqrt{2}(1+\frac{\pi }{4})$ and $y=\sqrt{2}(1-\frac{\pi }{4})$
Hence, equation of normal at $t=\frac{\pi }{4}$ is $x+y-2\sqrt{2}=0$
Distance of normal from the origin $=\left|\frac{-2\sqrt{2}}{\sqrt{1+1}}\right|=2$ units