Question

Show that the normal at any point θ to the curve $x=acosθ+aθsinθ,y=asinθ-aθcosθ$ is at a constant distance from the origin.

Answer 1

We have x=acosθ+aθsinθ
$∴\frac{dx}{dθ}=-asinθ+asinθ+aθcosθ=aθcosθ$
$y=asinθ-aθcosθ$
∴\frac{dy}{dθ}$$=acosθ-acosθ+aθsinθ=aθsinθ
∴ $\frac{dy}{dx}$=$\frac{dy}{dθ}$ . $\frac{dθ}{dx}$=$\frac{aθsinθ}{aθcosθ}$=tanθ

∴ Slope of the normal at any point θ is -$\frac{1}{tanθ}$.
The equation of the normal at a given point (x,y) is given by,
$y-asinθ+aθcosθ=$-\frac{1}{tanθ}$$(x-acosθ-aθsinθ)
$⇒ysinθ-asin^2θ+aθsinθcosθ=-xcosθ+acos^2θ+aθsinθcosθ$
$⇒xcosθ+ysinθ-a(sin^2θ+cos^2θ)=0$
$⇒xcosθ+ysinθ-a=0$
Now, the perpendicular distance of the normal from the origin is
$\frac{|-a|}{\sqrt{cos^2θ+sin^2θ}}$ $=\frac{|-a|}{\sqrt1}=|-a|,$ which is independent of θ.
Hence, the perpendicular distance of the normal from the origin is constant