Let RS be the diameter of the circle x² + y² =1, where S = (1, 0). Let P be a variable point (other than R and S) on the circle and tangents to the circle at ‘S’ and ‘P’ meet at the point ‘Q’. The normal to the circle at ‘P’ intersects a line drawn through ‘Q’ parallel to ‘RS’ at a point E then the locus of ‘E’. passes through the point(s) 

We can see the line on which point E lies is having equation y=tanθx now as we can see the y co-ordinate of E is $\frac{1-\cos \theta }{\sin \theta }$

subsituting in the given  equation we get the co-ordinate of E as

$(\frac{1-\cos \theta }{\sin \theta \tan \theta },\frac{1-\cos \theta }{\sin \theta })$

$\Rightarrow E=(\frac{\tan {\displaystyle \frac{\theta }{2}}}{\tan \theta },\tan \frac{\theta }{2})$=(h.k)

$$\begin{gathered} \Rightarrow k=\tan \frac{\theta }{2} and h=\frac{k}{\tan \theta } \\ \Rightarrow \frac{2\tan \frac{\theta }{2}}{1-{\tan }^2\frac{\theta }{2}}=\frac{k}{h} \\ \Rightarrow \frac{2k}{1-k^2}=\frac{k}{h} \\ \therefore locus is 2xy=y(1-y^2) \\ By verification options A and C satiesfies the locus \end{gathered}$$