An ellipse and a hyperbola have the same cent | vedclass

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(B) Let the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with eccentricity $e_1$. The foci are $(\pm ae_1, 0)$.Let the hyperbola be $\frac{x^2}{

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(B) Let the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with eccentricity $e_1$. The foci are $(\pm ae_1, 0)$.Let the hyperbola be $\frac{x^2}{A^2} - \frac{y^2}{B^2} = 1$ with eccentricity $e_2$. The foci are $(\pm Ae_2, 0)$.Since they have the same foci,$ae_1 = Ae_2$.Given that the minor axis of the ellipse $(2b)$ is equal to the conjugate axis of the hyperbola $(2B)$,we have $b = B$,so $b^2 = B^2$.For the ellipse,$b^2 = a^2(1 - e_1^2)$.For the hyperbola,$B^2 = A^2(e_2^2 - 1)$.Equating $b^2$ and $B^2$,we get $a^2(1 - e_1^2) = A^2(e_2^2 - 1)$.Since $ae_1 = Ae_2$,we have $A = \frac{ae_1}{e_2}$.Substituting $A$ into the equation: $a^2 - a^2e_1^2 = (\frac{ae_1}{e_2})^2(e_2^2 - 1) = a^2e_1^2 - \frac{a^2e_1^2}{e_2^2}$.Dividing by $a^2$: $1 - e_1^2 = e_1^2 - \frac{e_1^2}{e_2^2}$.$1 = 2e_1^2 - \frac{e_1^2}{e_2^2}$.Dividing by $e_1^2$: $\frac{1}{e_1^2} = 2 - \frac{1}{e_2^2}$.Therefore,$e_1^{-2} + e_2^{-2} = 2$.

An ellipse and a hyperbola have the same centre at the origin,the same foci,and the minor axis of the one is the same as the conjugate axis of the other. If $e_1$ and $e_2$ are their eccentricities respectively,then $e_1^{-2} + e_2^{-2}$ equals:

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