Parabola Part 2

Concept:Use the section formula to find coordinates of point dividing a segment, then eliminate parameter to get locus. For a chord bisected at a point, use the property $T = S_1$ for the conic.Explanation:Vertex of parabola $x^2 = 4y$ is $O(0,0)$.Take any point $Q$ on parabola as $(2t, t^2)$.Point $P$ divides $OQ$ internally in ratio $2:3$.Using section formula: $x = \frac{2(2t) + 3(0)}{5} = \frac{4t}{5}$, $y = \frac{2(t^2) + 3(0)}{5} = \frac{2t^2}{5}$.Eliminate $t$: $t = \frac{5x}{4}$, then $y = \frac{2}{5}\left(\frac{5x}{4}\right)^2 = \frac{5x^2}{8}$.Thus conic $C$: $5x^2 = 8y$ or $5x^2 - 8y = 0$.We need chord of $C$ bisected at $(1,2)$.For $S(x,y) = 5x^2 - 8y$, chord equation is $T = S_1$.Compute $S_1 = 5(1)^2 - 8(2) = 5 - 16 = -11$.Compute $T$: replace $x^2 \to x(1)$, $y \to \frac{y+2}{2}$. So $T = 5x - 8\left(\frac{y+2}{2}\right) = 5x - 4(y+2) = 5x - 4y - 8$.Set $T = S_1$: $5x - 4y - 8 = -11 \implies 5x - 4y + 3 = 0$.Answer:Option A: $5x - 4y + 3 = 0$