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An ellipse and a hyperbola are confocal ...

An ellipse and a hyperbola are confocal (have the same focus) and the conjugate axis of the hyperbola is equal to the minor axis of the ellipse. If `e_1a n de_2` are the eccentricities of the ellipse and the hyperbola, respectively, then prove that `1/(e_1^2)+1/(e_2^2)=2` .

Text Solution

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Let the foci of the ellipse be `(pmae_(1),0)` and those of the hyperbola be `(pmAe_(2),0)`

According to the question, we have
`ae_(1)=Ae_(2)" (1)"`
Also, it is given that the conjugate axis of the hyperbola is equals to the minor axis of the ellipse. Therefore,
`a^(2)(1-e_(1)^(2))=A^(2)(e_(2)^(2)-1)" (2)"`
From (1) and (2), we have
`(1)/(e_(1)^(2))+(1)/(e_(2)^(2))=2`
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