Locus of Intersections from Circle Tangents and Normals

To find the locus of the point $E$, we analyze the geometric conditions given in the problem step-by-step.

Step 1: Coordinates of points $S, R,$ and $P$

The given circle is $x^2 + y^2 = 1$ centered at the origin $O(0, 0)$.

• The point $S$ is given as $(1, 0)$.
• Since $RS$ is a diameter, $R$ is $(-1, 0)$, and the line $RS$ lies along the $x$-axis (i.e., $y = 0$).
• Let $P$ be a variable point on the circle represented in parametric form as $P(\cos\theta, \sin\theta)$, where $\theta \neq 0, \pi$.

Step 2: Finding the intersection point $Q$

• The tangent to the circle at $S(1, 0)$ is the vertical line: $x = 1$
• The tangent to the circle at $P(\cos\theta, \sin\theta)$ is: $x\cos\theta + y\sin\theta = 1$

To find the point $Q$, where these two tangents intersect, substitute $x = 1$ into the tangent equation at $P$: $\cos\theta + y\sin\theta = 1$ $y\sin\theta = 1 - \cos\theta$ $y = \frac{1 - \cos\theta}{\sin\theta} = \tan\left(\frac{\theta}{2}\right)$

Thus, the coordinates of $Q$ are: $Q = \left(1, \tan\left(\frac{\theta}{2}\right)\right)$

Step 3: Finding the locus of point $E$

• The line drawn through $Q$ parallel to $RS$ (which is the $x$-axis) is the horizontal line: $y = \tan\left(\frac{\theta}{2}\right)$

• The normal line to the circle at $P(\cos\theta, \sin\theta)$ passes through the origin $(0,0)$ and $P$, so its equation is: $y = (\tan\theta) x$

• Point $E(x, y)$ is the point of intersection of these two lines: $y = \tan\left(\frac{\theta}{2}\right)$ $x = \frac{y}{\tan\theta}$

Using the double-angle identity for tangent, $\tan\theta = \frac{2\tan\left(\frac{\theta}{2}\right)}{1 - \tan^2\left(\frac{\theta}{2}\right)}$, we substitute $y = \tan\left(\frac{\theta}{2}\right)$: $\tan\theta = \frac{2y}{1 - y^2}$

Substituting this expression for $\tan\theta$ into the equation for $x$: $x = \frac{y}{\left(\frac{2y}{1 - y^2}\right)} = \frac{1 - y^2}{2}$

Rearranging this gives the equation of the locus of $E$: $2x = 1 - y^2 \implies y^2 = 1 - 2x$

Step 4: Checking the given options

We check which of the given points lie on the parabola $y^2 = 1 - 2x$:

1. For Option A: $\left(\frac{1}{3}, \frac{1}{\sqrt{3}}\right)$ $\text{LHS} = y^2 = \left(\frac{1}{\sqrt{3}}\right)^2 = \frac{1}{3}$ $\text{RHS} = 1 - 2x = 1 - 2\left(\frac{1}{3}\right) = \frac{1}{3}$ Since $\text{LHS} = \text{RHS}$, Option A lies on the locus.

2. For Option B: $\left(\frac{1}{4}, \frac{1}{2}\right)$ $\text{LHS} = y^2 = \left(\frac{1}{2}\right)^2 = \frac{1}{4}$ $\text{RHS} = 1 - 2x = 1 - 2\left(\frac{1}{4}\right) = \frac{1}{2}$ Since $\text{LHS} \neq \text{RHS}$, Option B does not lie on the locus.

3. For Option C: $\left(\frac{1}{3}, -\frac{1}{\sqrt{3}}\right)$ $\text{LHS} = y^2 = \left(-\frac{1}{\sqrt{3}}\right)^2 = \frac{1}{3}$ $\text{RHS} = 1 - 2x = 1 - 2\left(\frac{1}{3}\right) = \frac{1}{3}$ Since $\text{LHS} = \text{RHS}$, Option C lies on the locus.

4. For Option D: $\left(\frac{1}{4}, -\frac{1}{2}\right)$ $\text{LHS} = y^2 = \left(-\frac{1}{2}\right)^2 = \frac{1}{4}$ $\text{RHS} = 1 - 2\left(\frac{1}{4}\right) = \frac{1}{2}$ Since $\text{LHS} \neq \text{RHS}$, Option D does not lie on the locus.

Thus, the locus passes through the points given in A and C.