Let RS be the diameter of the circle x ^ 2 + y ^ 2 = 1 where, S is...

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