The normal to the curve $x=a(\cos heta+ heta \sin heta), y=a(\sin heta-$ $ heta \cos heta)

Question

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

Professor PC Thomas Classes · Answer

(a, c): $\frac{d y}{d x}=\frac{d y}{d 	heta} \cdot \frac{d 	heta}{d x}$ $=	an 	heta=$ slope of tangent\ $	herefore \quad$ Slope of normal to the curve\ $=-\cot 	heta=	an (90+	heta)$.\ Now equation of normal to the curve\ $[y-a(\sin 	heta-	heta \cos 	heta)]$\ $=-\frac{\cos 	heta}{\sin 	heta}(x-a(\cos 	heta+	heta \sin 	heta))$\ $\Rightarrow x \cos 	heta+y \sin 	heta=a(1)$\ Now distance from $(0,0)$ to $x \cos 	heta+y \sin 	heta$ $=a$ is\ distance $(d)=\frac{|0+0-a|}{1}$\ $	herefore \quad$ Distance is constant $=a$.\