JEE Main 2026 Jan 28 Shift 2, Mathematics Q10: Introduction to Hyperbola

JEE Main2026Jan 28, Shift 2Mathematics

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 :

A
296
B
126
C
148
D
67

Solution

The ellipse has $a^{2} = 144, b^{2} = 169$ , so its major axis is along the $y$ -axis. Eccentricity of ellipse: $e^{'} = 1 - \frac{144}{169} = \frac{5}{13}$ . Foci of ellipse: $(0, \pm b e^{'}) = (0, \pm 5)$ .

The hyperbola $\frac{x ^{2}}{16} - \frac{y ^{2}}{λ ^{2}} = - 1$ rewrites as $\frac{y ^{2}}{λ ^{2}} - \frac{x ^{2}}{16} = 1$ — vertical transverse axis. Its foci: $(0, \pm λ 1 + 16/ λ^{2})$ .

Setting $λ 1 + 16/ λ^{2} = 5 \Rightarrow λ^{2} + 16 = 25 \Rightarrow λ = 3$ .

Eccentricity of $H$ : $e = 1 + \frac{16}{9} = \frac{5}{3}$ .

Length of latus rectum of $H$ $= \frac{2 \cdot 16}{3} = \frac{32}{3}$ .

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

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JEE Main 2026 Jan 28 Shift 2, Mathematics Q10: Introduction to Hyperbola

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