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