Parabola
Locus via Internal Division — Chord Bisected at Point
nta_pyq_2026_jan
Grade 11
Question:
Let $O$ be the vertex of the parabola $x^2=4y$ and $Q$ be any point on it. Let the locus of the point $P$, which divides the line segment $OQ$ internally in the ratio $2:3$ be the conic $C$. Then the equation of the chord of $C$, which is bisected at the point $(1,2)$, is:
5x-4y+3=0
5x-y-3=0
4x-5y+6=0
x-2y+3=0
Step-by-Step Solution
Key Concept: $Q=(2t,t^2)$ on $x^2=4y$. $P$ divides $OQ$ in $2:3$: $P=\left(\tfrac{4t}{5},\tfrac{2t^2}{5}\right)$. Let $(h,k)=P$: $h=\tfrac{4t}{5}$, $k=\tfrac{2t^2}{5}$. Eliminating $t$: $x^2=\tfrac{8y}{5}$, i.e., conic $C$ with $a=\tfrac{2}{5}$.
Chord: $5x-4y+3=0$.
Correct Answer: 1