
e1=1−10075=105=21e2=2 F1(6,1),F2(−4,1)2ae2=10⇒a=25⇒2a=5⇒α=54=1+a2b2⇒b2=3a2 b=3×25β=533α2+2β2=3×25+2×25×3=225
JEE Main 2024 — Mathematics Coordinate Geometry
Let the foci of a hyperbola H coincide with the foci of the ellipse E:100(x−1)2+75(y−1)2=1 and the eccentricity of the hyperbola H be the reciprocal of the eccentricity of the ellipse E. If the length of the transverse axis of H is α and the length of its conjugate axis is β, then 3α2+2β2 is equal to
Held on 9 Apr 2024 · Verified 6 Jul 2026.
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