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Question: Answered & Verified by Expert
Let H:x2a2-y2 b2=1,a>0, b>0, be a hyperbola such that the sum of lengths of the transverse and the conjugate axes is 422+14. If the eccentricity H is 112, then value of a2+b2 is equal to ______.
MathematicsHyperbolaJEE MainJEE Main 2022 (29 Jun Shift 1)
Solution:
2209 Upvotes Verified Answer
The correct answer is: 88

Given x2a2-y2b2=12a+2b=422+14 and e=112

We know  e2=1+b2a2114=1+b2a2b2=74a2

x2a2-y272a2=1 

Now 2a+2·7a2=422+14

a2+7=422+7

a=42a2=32

b2=74×16×2=56

Hence a2+b2=88

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