JEE Advanced · 2016 · 22 May 2016 · Paper 1 · Circle

Let $P(\cos \theta ,\sin \theta )$ be a point on the circle.

Tangent at $S(1,0)$: $x=1$.

Tangent at $P$: $x\cos \theta +y\sin \theta =1$.

These meet at $Q(1,\frac{1-\cos \theta }{\sin \theta })$.

Normal at $P$ passes through origin with slope $\frac{\sin \theta }{\cos \theta }$: equation $y=x\tan \theta$.

Line through $Q$ parallel to $RS$ (the $x$-axis): $y=\frac{1-\cos \theta }{\sin \theta }$.

Intersection $E$: $x\tan \theta =\frac{1-\cos \theta }{\sin \theta }$, so $x=\frac{1-\cos \theta }{\sin \theta }\cdot \frac{\cos \theta }{\sin \theta }$.

Using $\tan \frac{\theta }{2}=\frac{\sin \theta }{1+\cos \theta }=\frac{1-\cos \theta }{\sin \theta }$, let $\alpha =\tan \frac{\theta }{2}$.

Then $y=\alpha$ and from $\cos \theta +\alpha \sin \theta =1$, using half-angle formulas, we get the locus $y^2=1-2x$.

Checking: $(\frac{1}{3},\frac{1}{\sqrt{3}})$: $\frac{1}{3}=1-\frac{2}{3}=\frac{1}{3}$. True.

$(\frac{1}{3},-\frac{1}{\sqrt{3}})$: also satisfies. True.