Question

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

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

Answer 1

The correct option is C $2$

$x=2\cos t+2t\sin t,\frac{dx}{dt}=2t\cos ty=2\sin t–2t\cos t\frac{dy}{dt}=2t\sin t\Rightarrow \frac{dy}{dx}=\tan t$
So, slope of normal at $t=\frac{\pi }{4}$ is $-\frac{1}{\tan t}=-1$
Equation of normal at
$t=\frac{\pi }{4}x+y–2\sqrt{2}=0$
Distance of normal form the Origin= $\left|\frac{-2\sqrt{2}}{\sqrt{1+1}}\right|=2$ units