Given a2x2−b2y2=1, 2a+2b=4(22+14) and e=211
We know e2=1+a2b2⇒411=1+a2b2⇒b2=47a2
∴(a)2x2−(27a)2y2=1
Now 2a+2⋅27a=4(22+14)
a(2+7)=42(2+7)
a=42⇒a2=32
b2=47×16×2=56
Hence a2+b2=88
Let H:a2x2−b2y2=1,a>0,b>0, be a hyperbola such that the sum of lengths of the transverse and the conjugate axes is 4(22+14). If the eccentricity H is 211, then value of a2+b2 is equal to ______.
Held on 29 Jun 2022 · Verified 6 Jul 2026.
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