ParabolamediumPYQ · JEE Main · 4 Apr 2026 · Shift 1 (Morning)Free

Parabola-Ellipse Shared Latus Rectum: e²+2√2 = 3 | JEE 2026

JEE Maths question with a full step-by-step solution.

Question
Consider the parabola P:y2=4kxP:y^2=4kx and the ellipse E:x2a2+y2b2=1E:\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1. Let the line segment joining the points of intersection of PP and EE be their latus rectums. If the eccentricity of EE is ee, then e2+22e^2+2\sqrt2 is equal to
Solution
Answer: 3 (± 0.01)
Step 1: Focus of the parabola is (k,0)(k,0); focus of the ellipse is (ae,0)(ae,0). Since their latus rectums coincide, the foci coincide:
k=ae.(1)k=ae.\quad(1)
 Step 2: Latus rectum lengths equal: parabola =4k=4k, ellipse =2b2a=\dfrac{2b^2}{a}:
4k=2b2a ⇒ 4ae=2b2a ⇒ 4a2e=2b2.4k=\frac{2b^2}{a}\ \Rightarrow\ 4ae=\frac{2b^2}{a}\ \Rightarrow\ 4a^2e=2b^2.
 Step 3: Use b2=a2(1−e2)b^2=a^2(1-e^2):
4a2e=2a2(1−e2) ⇒ 2e=1−e2 ⇒ e2+2e−1=0 ⇒ (e+1)2=2 ⇒ e=2−1.4a^2e=2a^2(1-e^2)\ \Rightarrow\ 2e=1-e^2\ \Rightarrow\ e^2+2e-1=0\ \Rightarrow\ (e+1)^2=2\ \Rightarrow\ e=\sqrt2-1.
 Step 4:
e2+22=(2−1)2+22=(3−22)+22=3.e^2+2\sqrt2=(\sqrt2-1)^2+2\sqrt2=(3-2\sqrt2)+2\sqrt2=3.
 Correct answer: 3
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