Let … be the vertex of the parabola … and…

Single correct+4 / -1Answer matches NTA’s final key

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

Solution

Take $P(t_1^2, 2t_1)$ and $Q(t_2^2, 2t_2)$ . Slope of $OP$ is $\frac{2}{t_1}$ , of $OQ$ is $\frac{2}{t_2}$ ; perpendicular means $\frac{4}{t_1 t_2} = -1$ , so $t_1t_2 = -4$ .

Midpoint $(h, k)$ : $h = \frac{t_1^2 + t_2^2}{2}$ , $k = t_1 + t_2$ . Then

$k^2 = t_1^2 + t_2^2 + 2t_1t_2 = 2h - 8.$

The locus is $y^2 = 2(x - 4)$ , a parabola with $4a = 2$ .

Answer: latus rectum $= 2$ .

Answer checked against the official answer key. The solution is our own working.

JEE MAIN 2026 · Co-ordinate Geometry

Solution

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