🛠️ JEE➗ Maths

Let the ellipse \(E:\frac{{x}^{2}}{144}+\frac{{y}^{2}}{169}=1\) and the hyperbola \(H:\frac{{x}^{2}}{16}−\frac{{y}^{2}}{…

Q1

Let the ellipse \(E:\frac{{x}^{2}}{144}+\frac{{y}^{2}}{169}=1\) and the hyperbola \(H:\frac{{x}^{2}}{16}−\frac{{y}^{2}}{{\lambda }^{2}}=−1\) 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 \(24(e+L)\)is:

[JEE Main 2026, 28 Jan (Shift 2)]

a

126

b

148

c

67

d

296

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