Question

The normal to the curve $x=a(1+cos\theta \text{)},$ $y=a\sin \theta$ at $\theta$ always passes through the fixed point

• A. $(a,0)$
• B. $(0,a)$
• C. (0, 0)
• D. $(a,a)$

Answer 1

Answer: (A)

We have, $x=a(1+\cos \theta ),y=a\sin \theta$ $\frac{dx}{d\theta }=a(-\sin \theta ),\frac{dy}{d\theta }=a\cos \theta$ $\Rightarrow$ $\frac{dy}{dx}=\frac{dy/d\theta }{dx/d\theta }=-\frac{\cos \theta }{\sin \theta }$ $\therefore$ Equation of normal at $[a(1+\cos \theta ).a\sin \theta ]$ $(y-a\sin \theta )=\frac{\sin \theta }{\cos \theta }[x-a(1+\cos \theta )]$ It is clear that in the given options normal passes through the point (a, 0).