If normal to curve $x=3\cos\theta-\cos^3{\theta}$ and $y=3\sin\theta-\sin^3{\theta}$ at point $P(\theta)$ passes through the origin, then $\theta=$

Question

If normal to curve $x=3\cos\theta-\cos^3{\theta}$ and $y=3\sin\theta-\sin^3{\theta}$ at point $P(\theta)$ passes through the origin, then $\theta=$

(A) $0$

(B) $\dfrac{\pi}{4}$

(C) $\dfrac{\pi}{2}$

(D) $\dfrac{\pi}{6}$

My Approach:

I got the slope of normal as $\tan^3{\theta}.$

So equation of normal will be $$y-3\sin\theta +\sin^3{\theta}=\tan^3{\theta \cdot(x-3\cos\theta+3\cos^3{\theta})}$$

Because it passes through origin so I put $x=0,y=0$, which lead me to the equation,

$$3\sin\theta \cdot \cos{\theta} \cdot\cos{2\theta}=0$$

So value of $\theta$ will be $\theta=0, \dfrac{\pi}{2},\dfrac{\pi}{4}.$

But answer given is only $\theta = \dfrac{\pi}{4}$

My doubt:

Why option (A) and (C) are wrong.

Answer 1

Your equation should be: $$(y-3\sin{\theta}+\sin^3{\theta}) = \tan^3{\theta}(x-3\cos{\theta}+\cos^3{\theta})$$ When you rewrite $\tan^3{\theta}$ as $\frac{\sin^3{\theta}}{\cos^3{\theta}}$ and substitute $x=y=0$, you get $\tan^2{\theta}=1$ or $\sin{2\theta}=0$. Your explanation is right, $\theta=0,\frac{\pi}{2}$ work as well.