Question

The normal to the curve $x=a\left(\cos \left(\theta \right)+\sin \left(\theta \right)\right),y=a\left(\sin \left(\theta \right)-\theta \cos \left(\theta \right)\right)$ at any point $\theta$ is such that

• A. It is a constant distance from the origin
• B. It passes through $\left(a\frac{\pi }{2},-a\right)$
• C. It makes $\frac{\pi }{2}-\theta$ with the $x$-axis
• D. It passes through the origin

Answer 1

The correct option is A

It is a constant distance from the origin

Find the normal to the curve

We know that equation of normal is

$\left(y-y_0\right)=-\frac{1}{{\displaystyle \frac{dy}{dx}}}\left(x-x_0\right)$

$\frac{dy}{dx}=\tan \left(\theta \right)$

The equation of the normal at $\theta$ is

$$\begin{gathered} \left(y-a\left(\sin \left(\theta \right)-\theta \cos \left(\theta \right)\right)\right)=-\frac{1}{{\displaystyle \tan \left(\theta \right)}}\left(x-a\left(\cos \left(\theta \right)+\sin \left(\theta \right)\right)\right) \\ \Rightarrow y\sin \left(\theta \right)-a{\sin }^2\left(\theta \right)+a\sin \left(\theta \right)\cos \left(\theta \right)=-x\cos \left(\theta \right)+a{\cos }^2\left(\theta \right)+a\sin \left(\theta \right)\cos \left(\theta \right) \\ \Rightarrow y\sin \left(\theta \right)+x\cos \left(\theta \right)-a\left({\sin }^2\left(\theta \right)+{\cos }^2\left(\theta \right)\right)=0 \\ \Rightarrow y\sin \left(\theta \right)+x\cos \left(\theta \right)-a=0 \end{gathered}$$

It is an equation of straight line which is at a constant distance from the origin $\left(0,0\right)$

Hence, the correct option is $\left(\mathrm{A}\right)$.