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← JEE Main 2022 · 27 Jul · Shift 1
MathematicsInteger+4 / -1Question 24 of 90

JEE Main 2022 · 27 Jul · Shift 1, Mathematics question 24

An ellipse E:x2a2+y2b2=1E:\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1 passes through the vertices of the hyperbola H:x249−y264=−1H:\dfrac{x^2}{49}-\dfrac{y^2}{64}=-1. Let the major and minor axes of the ellipse E coincide with the transverse and conjugate axes of the hyperbola H, respectively. Let the product of the eccentricities of E and H be 12\dfrac12. If ll is the length of the latus rectum of the ellipse E, then the value of 113l113l is equal to ______.

Answer and worked solution

Official answer: 1552

Hyperbola. Vertices (0,±8)(0,\pm8) and eH2=1+4964=11364e_H^2=1+\frac{49}{64}=\frac{113}{64}.

Ellipse. Major axis along y with semi-axis 8; eE2=1/4113/64=16113e_E^2=\frac{1/4}{113/64}=\frac{16}{113}, so a2=64⋅97113a^2=64\cdot\frac{97}{113}.

Latus rectum l=2a28=16⋅97113l=\frac{2a^2}8=\frac{16\cdot97}{113}, so 113l=1552113l=1552.

Answer: 1552