JEE Main · 2026 · 04 Apr 2026 · Shift 1 · Ellipse

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 ____.

Answer: 3

Solution

The common chord is a latus rectum of both curves: for

y^{2} = 4 k x

it is

x = k

with ends

(k, \pm 2 k)

, and for the ellipse it is

x = a e

with ends

(a e, \pm b^{2} / a)

. Matching the chords gives

k = a e

and

2 k = \frac{b ^{2}}{a}

. Eliminating

k

2 a e = \frac{b ^{2}}{a} = a (1 - e^{2})

, so

e^{2} + 2 e - 1 = 0

and

e = 2 - 1

. Then

e^{2} = 3 - 22

, and

e^{2} + 22 = 3

JEE Main · 2026 · 04 Apr 2026 · Shift 1 · Integer answer