JEE Main 2026 — Apr 4, Shift 1 — Mathematics solution

JEE Main2026Apr 4, Shift 1Mathematics

Consider the parabola $P : y^{2} = 4 k x$ and the ellipse $E : \frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$ . Let the line segment joining the points of intersection of $P$ and $E$ , be their latus rectums. If the eccentricity of $E$ is $e$ , then $e^{2} + 22$ is equal to ______.

Your answer

Solution

The parabola's latus rectum is the vertical chord through its focus $(k, 0)$ , of length $4 k$ . The ellipse's latus rectum is the vertical chord through its focus $(a e, 0)$ , of length $\frac{2 b ^{2}}{a}$ .

For the same segment to serve as latus rectum for both, the foci coincide and the lengths match:

$k = a e and 4 k = \frac{2 b ^{2}}{a} .$

Substitute: $4 a e = \frac{2 b ^{2}}{a} \Rightarrow 2 a^{2} e = b^{2} = a^{2} (1 - e^{2})$ , giving

$2 e = 1 - e^{2} \Rightarrow e^{2} + 2 e - 1 = 0.$

$\Rightarrow (e + 1)^{2} = 2 \Rightarrow e = 2 - 1$ (taking $e > 0$ ).

Then $e^{2} = 3 - 22$ , so

$e^{2} + 22 = 3.$

← Back to full paper

Practice more Mathematics

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →