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ABCD is a rhombus with AC=2BD. Diagonal...

ABCD is a rhombus with AC=2BD. Diagonals AC and BD intersect at `P.E_(1),E_(2),E_(3) and E_(4)` are four ellipes 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,e_(i)` are the eccentricities fo `E_(i)`, then

A

`e_(i)=e_(3)`

B

`e_(2)=e_(4)`

C

`e_(1)=2e_(2)`

D

`e_(1)=e_(2)`

Text Solution

Verified by Experts


AB=BC=CD=DA=
Since `DeltaPAB, DeltaPBC, DeltaPCD, DeltaPDA` are congruent,
`e_(1)=e_(2)=e_(3)=3_(4)=(a)/(PA+PB)`
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