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) $(1) (\frac{1}{3}, 1/\sqrt{3}) (2) (\frac{1}{4}, \frac{1}{2}) (3) (\frac{1}{3}, -1/\sqrt{3}) (4) (\frac{1}{4}, -\frac{1}{2})$

Let RS be the diameter of the circle x² + y² = 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) $x^{2}+y^{2}=1, \quad \text{centre } O(0,0), \ r=1$ $S(1,0), \quad R(-1,0), \quad RS \text{ is a diameter}$ $P \text{ variable on the circle}$ $Q = \text{meeting of tangents at } S \text{ and } P$ $E = \text{normal at } P \ \cap \ \text{line through } Q \parallel RS$ $\text{To find: the locus of } E$ $x^2+y^2=1, \quad O(0,0), \ r=1$ $R(-1,0), \ S(1,0), \quad RS \text{ on the } x\text{-axis}$ $P=(\cos\theta,\ \sin\theta)$ $\angle SOP=\theta, \quad OP \text{ is the radius}$ $\text{tangent at } S(1,0):\ x=1$ $\text{tangent at } P:\ x\cos\theta+y\sin\theta=1$ $Q=(1,y_Q), \quad y_Q=\dfrac{1-\cos\theta}{\sin\theta}=\tan\dfrac{\theta}{2}$ $\text{normal at } P = \text{line } OP:\ y=x\tan\theta$ $E=\lambda(\cos\theta,\ \sin\theta)$ $\text{line through } Q \parallel RS:\ y=\tan\dfrac{\theta}{2}$ $E_y=\tan\dfrac{\theta}{2}$ $\lambda\sin\theta=\tan\dfrac{\theta}{2}=\dfrac{\sin\theta}{1+\cos\theta}$ $\lambda=\dfrac{1}{1+\cos\theta}$ $E=\left(\dfrac{\cos\theta}{1+\cos\theta},\ \dfrac{\sin\theta}{1+\cos\theta}\right)$ $X=\dfrac{\cos\theta}{1+\cos\theta} \Rightarrow \cos\theta=\dfrac{X}{1-X}$ $Y=\dfrac{\sin\theta}{1+\cos\theta} \Rightarrow \sin\theta=\dfrac{Y}{1-X}$ $\cos^2\theta+\sin^2\theta=1 \Rightarrow \dfrac{X^2+Y^2}{(1-X)^2}=1$ $X^2+Y^2=(1-X)^2$ $Y^2=1-2X$ $y^2=1-2x$ $y^2=-2\left(x-\dfrac{1}{2}\right)$ $\text{parabola, vertex } \left(\dfrac{1}{2},0\right), \text{ opening left}$ $\left(\dfrac{1}{3},\pm\dfrac{1}{\sqrt3}\right):\ 1-2\cdot\dfrac{1}{3}=\dfrac{1}{3}=\left(\dfrac{1}{\sqrt3}\right)^2\ \checkmark$ $\text{Options (1) and (3) lie on the locus}$ $\left(\dfrac{1}{4},\pm\dfrac{1}{2}\right):\ 1-2\cdot\dfrac{1}{4}=\dfrac{1}{2}\ne\dfrac{1}{4}\ \times$ $\text{Options (2) and (4) do not lie on the locus}$ $y^2=1-2x$ $\left(\dfrac{1}{3},\pm\dfrac{1}{\sqrt3}\right) \text{ lie on it}$ $\boxed{\text{Answer: (1) and (3)}}$