Let the ellipse \(\mathrm{E}: \frac{\mathrm{x}^{2}}{144}+\frac{\mathrm{y}^{2}}{169}=1\) and the hyperbola \(\mathrm{H}: \frac{\mathrm{x}^{2}}{16}-\frac{\mathrm{y}^{2}}{\lambda^{2}}=-1\) have the same foci. If e and L respectively denote the eccentricity and the length of the latus rectum of H , then the value of \(24(\mathrm{e}+\mathrm{L})\) is :

JEE Main · 2026 · 28 Jan 2026 · Shift 2 · Hyperbola

Let the ellipse

E : \frac{x ^{2}}{144} + \frac{y ^{2}}{169} = 1

and the hyperbola

H : \frac{x ^{2}}{16} - \frac{y ^{2}}{λ ^{2}} = - 1

have the same foci. If e and L respectively denote the eccentricity and the length of the latus rectum of H , then the value of

24 (e + L)

is :

Answer: (A)

For the ellipse

\frac{x ^{2}}{144} + \frac{y ^{2}}{169} = 1

the larger denominator sits under

y^{2}

, so

a^{2} = 169, b^{2} = 144

and

c^{2} = 169 - 144 = 25

, giving foci

(0, \pm 5)

. The hyperbola

\frac{x ^{2}}{16} - \frac{y ^{2}}{λ ^{2}} = - 1

\frac{y ^{2}}{λ ^{2}} - \frac{x ^{2}}{16} = 1

, with transverse axis along

y

, so

a^{2} = λ^{2}, b^{2} = 16

and

c^{2} = λ^{2} + 16

. Equal foci require

λ^{2} + 16 = 25

, hence

λ^{2} = 9, λ = 3

. The eccentricity is

e = \frac{c}{a} = \frac{5}{3}

and the latus rectum is

L = \frac{2 b ^{2}}{a} = \frac{32}{3}

. Therefore

24 (e + L) = 24 (\frac{5}{3} + \frac{32}{3}) = 24 \cdot \frac{37}{3} = 296

JEE Main · 2026 · 28 Jan 2026 · Shift 2 · Single correct option