Let the length of major axis of an ellipse equal to the length of transverse axis of a hyperbola and their director circle have radius equal to 2R and R respectively. If e1 and e2 are the eccentricities of the ellipse and hyperbola then correct relation is

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4e12−2e22=6

e12−4e22=6

2e12−e22=6

4e22−e12=6

answer is D.

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Detailed Solution

x2a2+y2b2=1 andx2a2−y2B2=1 a2+b2=4R2 =a2(2− e12) anda2−B2=R2=a2(2− e22)4=2− e1 22− e2 2 ⇒ 4e2 2− e12=6

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