Question

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

• A. it passes through origin
• B. it passes through the point $(1,1)$
• C. it passes through $(\frac{a\pi }{2},-a)$
• D. it is at a constant distance from the origin

Answer 1

The correct option is A it passes through origin

$x=a\cos \theta +a\theta \sin \theta$

$\frac{dx}{d\theta }=-a\sin \theta +a\left\{\theta \right(-\cos \theta )+\sin \theta \}$

$=-a\theta \cos \theta$

$y=a(\sin \theta -\theta \cos \theta )$

$\frac{dy}{dx}=a\cos \theta -a\{\theta \sin \theta +\cos \theta \}$

$\therefore \frac{dy}{dx}=\frac{-a\theta \sin \theta }{-a\theta \cos \theta }=\tan \theta$

The equation of normal at $\theta$ is:

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

or, $y-a(\sin \theta -\theta \cos \theta )=-\frac{\cos \theta }{\sin \theta }(x-a\cos \theta -a\theta \sin \theta )$

or, $y\sin \theta -{\sin }^2\theta +a\theta \sin \theta \cos \theta$

$=-x\cos \theta +a{\cos }^2\theta +a\theta \sin \theta \cos \theta$

or, $y\sin \theta +x\cos \theta =a$

which wakes it clear that normal passes through origin $(0,0)$