ABCD is a rhombus with AC = 2BD. Diagonals AC and BD intersect at P. E1, E2, E3 and E4 arc four ellipses passing through P and their foci are A and B, B and C, C and D and D and A, respectively. If for i =1,2,3,4, ei are the eccentricities of Ei then

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e1 = e3

e2 = e4

e1 = 2e2

e1 = e2

answer is A.

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Detailed Solution

AB = BC = CD = DA = aSince, ΔPAB,△PBC,ΔPCD,△PDA are congruent,e1=e2=e3=e4=aPA+PB

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