The normal to the curve x = a(1 + cos [ ]), [y = a ] at [' '] always passes through the fi…

Question

The normal to the curve x = a(1 + cos $$\theta $$), $$y = a\sin \theta $$ at $$'\theta '$$ always passes through the fixed point

Accepted Answer

$${{dx} \over {d\theta }} =  - a\sin \theta $$ and $${{dy} \over {d\theta }} = a\cos \theta $$

$$\therefore$$ $${{dy} \over {dx}} =  - \cot \theta .$$

$$\therefore$$ The slope of the normal at $$\theta  $$ = $$ - {1 \over { - \cot \theta }}$$ $= \tan \theta $$\therefore$The equation of the normal at$\theta $is$y - a\sin \theta  = \tan \theta \left( {x - a - a\cos \theta } \right) (

\] \Rightarrow $$y - a\sin \theta  =$${{\sin \theta } \over {\cos \theta }}$$\left( {x - a - a\cos \theta } \right) (

\] \Rightarrow y\cos \theta  - a\sin \theta \cos \theta $$\,\,\,\,\,\,\,\,\,\,\,$$ = x\sin \theta  - a\sin \theta  - a\sin \theta \cos \theta $$ \Rightarrow x\sin \theta  - y\cos \theta  = a\sin \theta $$ \Rightarrow y = \left( {x - a} \right)\tan \theta \[

which always passes through (a, 0)$

Solutions · 1