Let RS be the diameter of the circle  $x^2+y^2=1$, where S is the point $(1, 0).$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 point E. Then the locus of E passes through the point(s)  

Circle:  $x^2+y^2=1$
    Equation of tangent at  $P(\cos \theta ,\sin \theta )$
                              $x\cos \theta +y\sin \theta =1$
    Equation of normal at P
                          $y=x\tan \theta$
   Equation of tangent at S is  $x=1$
 $\therefore Q(1,\frac{1-\cos \theta }{\sin \theta })=Q(1,\tan \frac{\theta }{2})$
  
 $\therefore$  Equation of the through Q and parallel to RS is  $y=\tan \frac{\theta }{2}$
$\therefore$   Intersection point E of normal and  $y=\tan \frac{\theta }{2}$
   $\tan \frac{\theta }{2}=x\tan \theta$
 $\Rightarrow x=\frac{1-{\tan }^2\frac{\theta }{2}}{2}$
$\therefore$  Locus of  $E:x=\frac{1-y^2}{2}or{y}^2=1-2x$
  It is satisfied by the points  $(\frac{1}{3},\frac{1}{\sqrt{3}})and(\frac{1}{3},\frac{-1}{\sqrt{3}})$