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JEE Main 202628 JanuaryEvening Shift
Exam
JEE Main
Year
2026
Date
28 January
Shift
Evening Shift
Subject
Mathematics
Chapter
Ellipse

JEE Main 2026 · 28 January · Evening Shift · Mathematics · Ellipse

Let the ellipse E: x^ 2 144 + y^ 2 169 =1 and the hyperbola H: x^ 2...

Let the ellipse E:x2144+y2169=1\mathrm{E}: \frac{x^{2}}{144}+\frac{y^{2}}{169}=1 and the hyperbola H:x216−y2λ2=−1\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(e+L)24(\mathrm{e}+\mathrm{L}) is :

Options

  1. A148
  2. B126
  3. C67
  4. D296

Correct answer

D. 296

Step-by-step solution

Ellipse E: x2144+y2169=1\frac{x^2}{144} + \frac{y^2}{169} = 1 has a2=169,b2=144a^2 = 169, b^2 = 144, so cE=25=5c_E = \sqrt{25} = 5 with foci at (0,±5)(0, \pm 5).
Hyperbola H: y2x2−x216=1\frac{y^2}{x^2} - \frac{x^2}{16} = 1 has the same foci, so cH2=x2+16=25c_H^2 = x^2 + 16 = 25, giving x2=9x^2 = 9.
Thus H is y29−x216=1\frac{y^2}{9} - \frac{x^2}{16} = 1 with a=3,b=4,c=5a = 3, b = 4, c = 5.
Eccentricity: e=53e = \frac{5}{3}.
Latus rectum: L=2b2a=323L = \frac{2b^2}{a} = \frac{32}{3}.
Therefore 24(e+L)=24(53+323)=24⋅373=29624(e + L) = 24\left(\frac{5}{3} + \frac{32}{3}\right) = 24 \cdot \frac{37}{3} = 296
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