Let RS be the diameter of the circle x 2 + y 2 = 1 where, S is the point ( 1 ,   0 ) . Let P   be…

1. $\left(\frac{1}{3},\frac{1}{\sqrt{3}}\right)$
2. $\left(\frac{1}{4},\frac{1}{2}\right)$
3. $\left(\frac{1}{3}, -\frac{1}{\sqrt{3}}\right)$
4. $\left(\frac{1}{4}, -\frac{1}{2}\right)$

Solution

$\mathrm{Let\; P\; be }(\mathrm{cos\; \theta },\mathrm{sin\; \theta })$, where
$\mathrm{\theta }\ne 0,\mathrm{\pi }$
Tangent at $P :x\cos \theta +y\sin \theta =1$ ........(i)
Tangent at $S :x=1$ .........(ii)
$\therefore$ By (i) and (ii) : $Q\left(1,\frac{1-\cos \theta }{\sin \theta }\right)$
Line through Q parallel to RS :
$y=\frac{1-\cos \theta }{\sin \theta } \Rightarrow y=\tan \frac{\theta }{2}$ ..........(iii)
Normal at $P :y=\frac{\sin \theta }{\cos \theta }x \Rightarrow y=\tan \theta .x$ $\Rightarrow y=\left(\frac{2t\mathit{an}\frac{\theta }{2}}{1-t{\mathit{an}}^{\mathit{2}}\frac{\theta }{2}}\right)x$ ........(iv)  
Point of intersection of equation (iii) and (iv), $E :h=\frac{1-{\tan }^2\frac{\theta }{2}}{2};k=\tan \frac{\theta }{2}$
Eliminating $\theta :h=\frac{1-k^2}{2} \Rightarrow y^2=1-2x$
$\left(\frac{1}{3},\frac{1}{\sqrt{3}}\right) \& \left(\frac{1}{3},-\frac{1}{\sqrt{3}}\right)$  satisfy the locus.

Asked in: JEE Advanced 2016 (Paper 1)

Practice more Hyperbola questions on Aicharya