Q40 / 75 · JEE Main 2026 (28 January Shift 2) · Mathematics
Let the ellipse $\mathrm{E}: \frac{x^{2}}{144}+\frac{y^{2}}{169}=1$ and the hyperbola $\mathrm{H}: \frac{x^{2}}{16}-\frac{y^{2}}{\lambda^{2}}=-1$ 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 :
- A148
- B126
- C67
- Correct answerD296
Solution
Ellipse E: $\frac{x^2}{144} + \frac{y^2}{169} = 1$ has $a^2 = 169, b^2 = 144$, so $c_E = \sqrt{25} = 5$ with foci at $(0, \pm 5)$.
Hyperbola H: $\frac{y^2}{x^2} - \frac{x^2}{16} = 1$ has the same foci, so $c_H^2 = x^2 + 16 = 25$, giving $x^2 = 9$.
Thus H is $\frac{y^2}{9} - \frac{x^2}{16} = 1$ with $a = 3, b = 4, c = 5$.
Eccentricity: $e = \frac{5}{3}$.
Latus rectum: $L = \frac{2b^2}{a} = \frac{32}{3}$.
Therefore $24(e + L) = 24\left(\frac{5}{3} + \frac{32}{3}\right) = 24 \cdot \frac{37}{3} = 296$
Hyperbola H: $\frac{y^2}{x^2} - \frac{x^2}{16} = 1$ has the same foci, so $c_H^2 = x^2 + 16 = 25$, giving $x^2 = 9$.
Thus H is $\frac{y^2}{9} - \frac{x^2}{16} = 1$ with $a = 3, b = 4, c = 5$.
Eccentricity: $e = \frac{5}{3}$.
Latus rectum: $L = \frac{2b^2}{a} = \frac{32}{3}$.
Therefore $24(e + L) = 24\left(\frac{5}{3} + \frac{32}{3}\right) = 24 \cdot \frac{37}{3} = 296$
Hence, the correct answer is option: 'D'.