JEE Main 2026 Apr 8 Shift 2, Mathematics Q11: Parabola: Chord, Tangent and Normal

JEE Main2026Apr 8, Shift 2Mathematics

Let $O$ be the vertex of the parabola $y^{2} = 4 x$ and its chords $O P$ and $O Q$ are perpendicular to each other. If the locus of the mid-point of the line segment $P Q$ is a conic $C$ , then the length of its latus rectum is :

A
$1$
B
$2$
C
$4$
D
$8$

Solution

Parametrize: $P = (t_{1}^{2}, 2 t_{1})$ , $Q = (t_{2}^{2}, 2 t_{2})$ on $y^{2} = 4 x$ .

Slopes from origin: $\frac{2}{t _{1}}$ and $\frac{2}{t _{2}}$ . Perpendicularity gives $\frac{4}{t _{1} t _{2}} = - 1$ , hence $t_{1} t_{2} = - 4$ .

Midpoint $(h, k)$ of $P Q$ : $2 h = t_{1}^{2} + t_{2}^{2}$ and $2 k = 2 (t_{1} + t_{2})$ , so $t_{1} + t_{2} = k$ .

Using $(t_{1} + t_{2})^{2} = t_{1}^{2} + t_{2}^{2} + 2 t_{1} t_{2}$ :

$k^{2} = (t_{1}^{2} + t_{2}^{2}) + 2 t_{1} t_{2} = 2 h + 2 (- 4) = 2 h - 8$ .

So the locus is $y^{2} = 2 x - 8 = 2 (x - 4)$ $\to$ a parabola with $4 a = 2$ .

$∴$ Length of latus rectum $= 2$ .

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JEE Main 2026 Apr 8 Shift 2, Mathematics Q11: Parabola: Chord, Tangent and Normal

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