For ellipse 36x2+16y2=1: a=6, b=4, c=36−16=25.
Foci of ellipse: (±25,0).
For hyperbola with eccentricity e=5 and foci at (±25,0):
e=AC⇒A=525.
C2=A2+B2⇒20=54+B2⇒B2=596.
Latus rectum =A2B2=5252⋅596=596.
Let the foci of a hyperbola coincide with the foci of the ellipse 36x2+16y2=1. If the eccentricity of the hyperbola is 5, then the length of its latus rectum is :
Held on 21 Jan 2026 · Verified 6 Jul 2026.
16
245
12
596
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