Question

Show that the normal at any point $t$ to the curve $x=cost+atsint$ and $ysint-atcost$ is at constant distance from origin.

Answer 1

Consider the problem

we have the curve $x=acost+atsint$ and $y=asint-atcost$

Since, the slope of the normal is $-\frac{1}{\frac{dy}{dx}}$

So, first we will find $\frac{dy}{dx}$

$\begin{array}{c}\frac{dy}{dt}=a\cos t-a\left(\cos t-t\sin t\right)=at\sin t\\ \frac{dx}{dt}=-a\sin t+a\left(\sin t+t\cos t\right)=at\cos t\\ \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dy}{dx}}=\frac{at\sin t}{at\cos t}=\tan t\end{array}$

Equation of the normal at a point $\left(x_1,y_1\right)$ to any curve is given by

$y-y_1=-\frac{1}{\frac{dy}{dx}}\left(x-x_1\right)$

Therefore, the equation of the normal at point $t$ to the given curve is

$\begin{array}{c}y-\left(a\sin t-at\cos t\right)=-\frac{1}{\tan t}\left(x-a\cos t-at\sin t\right)\\ y-a\sin t+at\cos t=-x\cot t+a\frac{{\cos }^{2}t}{\sin t}+at\cos t\\ x\cot t+y-a\left(\sin t+\frac{{\cos }^{2}t}{\sin t}\right)=0\\ x\cot t+y-a\left(\frac{{\sin }^{2}t+{\cos }^{2}t}{\sin t}\right)=0\\ x\cot t+y-a\cos ect=0....\left(1\right)\end{array}$

So, equation $\left(1\right)$ is the equation of the normal to the curve

And, distance of the normal from the origin $\left(0,0\right)$ is

$\left|\frac{a\cos ect}{\sqrt{{\cot }^{2}t+1}}\right|=\left|\frac{a\cos ect}{\cos ect}\right|=\left|a\right|units$

which is independent of $t$ and the distance is constant from origin.