Question
The normal to the curve $$x = a\left( {\cos \theta+\theta \sin \theta } \right),\,y = a\left( {\sin \theta -\theta \cos \theta } \right)$$ at any point $$'\theta '$$ is such that
A.
it passes through the origin
B.
it makes an angle $$\frac{\pi }{2} + \theta $$ with the $$x - $$axis
C.
it passes through $$\left( {a\frac{\pi }{2}, - a} \right)$$
D.
it is at a constant distance from the origin
Answer :
it is at a constant distance from the origin
Solution :
$$\eqalign{
& x = a\left( {\cos \theta + \theta \sin \theta } \right) \cr
& \Rightarrow \frac{{dx}}{{d\theta }} = a\left( { - \sin \theta + \sin \theta + \theta \cos \theta } \right) \cr
& \Rightarrow \frac{{dx}}{{d\theta }} = a\theta \cos \theta \,......\left( 1 \right) \cr
& y = a\left( {\sin \theta - \theta \cos \theta } \right) \cr
& \frac{{dy}}{{d\theta }} = a\left[ {\cos \theta - \cos \theta + \theta \sin \theta } \right] \cr
& \Rightarrow \frac{{dy}}{{d\theta }} = a\theta \sin \theta \,......\left( 2 \right) \cr
& {\text{From equations}}\,\left( {\text{1}} \right)\,{\text{and}}\,\left( {\text{2}} \right){\text{, we get}} \cr
& \frac{{dy}}{{dx}} = \tan \theta \Rightarrow {\text{Slope of normal}} = - \cot \theta \cr
& {\text{Equation}}\,{\text{of}}\,{\text{normal}}\,{\text{at}}\,'\theta '\,{\text{is}}\,y - a\left( {\sin \theta - \theta \cos \theta } \right) \cr
& = - \cot \theta \left( {x - a} \right.\left( {\cos \theta + \theta \sin \theta } \right) \cr
& \Rightarrow y\sin \theta - a{\sin ^2}\theta + a\theta \cos \theta \sin \theta \cr
& = - x\cos \theta + a{\cos ^2}\theta + a\theta \sin \theta \cos \theta \cr
& \Rightarrow x\cos \theta + y\sin \theta = a \cr} $$
Clearly this is an equation of straight line which is at a constant distance $$'a'$$ from origin.