Parabola Part 2

Concept:The midpoint of a chord of a parabola satisfies the chord equation $T = S_1$, and internal division gives coordinates using section formula.Explanation:Let $M(x_1, y_1)$ be the midpoint of a chord of the parabola $y^2 = 4x$.The equation of the chord with midpoint $(x_1, y_1)$ is $y y_1 - 2(x + x_1) = y_1^2 - 4x_1$.Since the chord passes through $O(0,0)$, substitute $(0,0)$: $0 - 2(0 + x_1) = y_1^2 - 4x_1$ gives $-2x_1 = y_1^2 - 4x_1$.This simplifies to $y_1^2 = 2x_1$.Thus the locus $S$ is $y^2 = 2x$.Take any point $P$ on $S$: parametric form $P\left(\frac{1}{2}t^2, t\right)$.Let $Q$ divide $OP$ internally in the ratio $3:1$ (i.e., $OQ:QP = 3:1$).By section formula, $Q(x,y) = \left( \frac{3 \cdot \frac{t^2}{2} + 1 \cdot 0}{4},\; \frac{3 \cdot t + 1 \cdot 0}{4} \right) = \left( \frac{3t^2}{8},\; \frac{3t}{4} \right)$.Eliminate $t$: $t = \frac{4y}{3}$, so $t^2 = \frac{16y^2}{9}$.Substitute into $x = \frac{3t^2}{8}$: $x = \frac{3}{8} \cdot \frac{16y^2}{9} = \frac{2y^2}{3}$.Hence $3x = 2y^2$, i.e., $2y^2 = 3x$.Answer:The required locus is $2y^2 = 3x$, which corresponds to option B.