Question
The normal to the curve $$x = a\left( {1 + \cos \theta } \right),y = a\,\sin \theta $$ at $$'\theta '$$ always passes through the fixed point
A.
$$\left( {a,a} \right)$$
B.
$$\left( {0,a} \right)$$
C.
$$\left( {0,0} \right)$$
D.
$$\left( {a,0} \right)$$
Answer :
$$\left( {a,0} \right)$$
Solution :
$$\eqalign{
& \frac{{dx}}{{d\theta }} = - a\sin \theta \,{\text{and}}\,\frac{{dy}}{{d\theta }} = a\cos \theta \cr
& \therefore \frac{{dy}}{{dx}} = - \cot \theta \cr
& \therefore {\text{The slope of the normal}}\,{\text{at}}\,\theta = \tan \theta \cr
& \therefore {\text{The equation of the normal at}}\,\theta \,{\text{is}} \cr
& y - a\sin \theta = \tan \theta \left( {x - a - a\cos \theta } \right) \cr
& \Rightarrow y\cos \theta - a\sin \theta \cos \theta = x\sin \theta - a\sin \theta - a\sin \theta \cos \theta \cr
& \Rightarrow x\sin \theta - y\cos \theta = a\sin \theta \cr
& \Rightarrow y = \left( {x - a} \right)\tan \theta \cr
& {\text{which always passes through}}\,\left( {a,0} \right) \cr} $$