JEE Main · 2026 · 21 Jan 2026 · Shift 1 · Parabola

Let O be the vertex of the parabola

x^{2} = 4 y

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:

(A) $5 x - y - 3 = 0$
(B) $4 x - 5 y + 6 = 0$
(C) $x - 2 y + 3 = 0$
(D) $5 x - 4 y + 3 = 0$ ✓ Correct

Answer: (D)

Solution

A point

Q = (2 t, t^{2})

lies on

x^{2} = 4 y

, and

P

dividing

O Q

internally in

2 : 3

(\frac{4 t}{5}, \frac{2 t ^{2}}{5})

; eliminating

t

gives the locus

5 x^{2} = 8 y

. For the chord of this conic bisected at

(1, 2)

, the relation

T = S_{1}

reads

5 x - 4 (y + 2) = 5 - 16

, which simplifies to

5 x - 4 y + 3 = 0

JEE Main · 2026 · 21 Jan 2026 · Shift 1 · Single correct option

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