Consider the parabola P:y 2 = 4kx 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 rectum. Then e 2 + 2√(2) is equal to .

Mathematics · Ellipse

JEE Main 2026 — 4 April, Morning Shift — Question 44

Consider the parabola $\mathrm{P:y^2 = 4kx}$ and the ellipse $\mathrm{E:}\frac{\mathrm{x}^2}{\mathrm{a}^2} +\frac{\mathrm{y}^2}{\mathrm{b}^2} = 1$ . Let the line segment joining the points of intersection of P and E be their latus rectum. Then $\mathrm{e}^2 + 2\sqrt{2}$ is equal to ______.

Answer: 3

Numerical answer — enter this value.

Step-by-step solution

Focus of parabola ( $\mathrm{k}, 0$ ) Focus of Ellipse (ae, 0)

\mathrm{k}=\mathrm{ae} \end{gathered}$$ Also $4 \mathrm{k}=2 \frac{\mathrm{~b}^{2}}{\mathrm{a}}$ $\_\_\_\_$ $$\begin{gathered} 4 \mathrm{ae}=\frac{2 \mathrm{~b}^{2}}{\mathrm{a}} \end{gathered}$$ $\Rightarrow 4 \mathrm{a}^{2} \mathrm{e}=2 \mathrm{~b}^{2}$ $4 \mathrm{a}^{2} \mathrm{e}=2 \mathrm{a}^{2}\left(1-\mathrm{e}^{2}\right)$ $2 \mathrm{e}=1-\mathrm{e}^{2}$ $\mathrm{e}^{2}+2 \mathrm{e}+1=2$ $(\mathrm{e}+1)^{2}=2$ $\mathrm{e}+1=\sqrt{2} \therefore \mathrm{e}=\sqrt{2}-1$ $\mathrm{e}^{2}+2 \sqrt{2}=3$

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Exam: JEE Main 2026
Paper: 4 April, Morning Shift
Subject: Mathematics
Chapter: Ellipse
Topic: Special properties of ellipse