JEE Main 2026 · 4 Apr · Shift 1 · Mathematics Q22

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MathematicsInteger+4 / -1Question 22 of 75

Consider the parabola $P:y^2=4kx$ and the ellipse $E:\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ . 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 $e^2+2\sqrt2$ is equal to ________.

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Answer and worked solution

Official answer: 3

Common latus rectum. The parabola's latus rectum ends at $(k,\pm2k)$ ; the ellipse's at $\left(ae,\pm\frac{b^2}a\right)$ .

Match. $k=ae$ and $2ae=\dfrac{b^2}a=a(1-e^2)$ , so $e^2+2e-1=0$ and $e=\sqrt2-1$ .

Value. $e^2=3-2\sqrt2$ , so $e^2+2\sqrt2=3$ .

Answer: 3

Topics:Ellipse Parabola

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