Answer :

Given: e1 and e2 are respectively the eccentricities of the ellipse and the hyperbola


To find: value of 2e12 + e22




Eccentricity(e) of hyperbola is given by,



Here a = 3 and b = 2





Therefore,



For ellipse:




Eccentricity(e) of ellipse is given by,







Therefore,





Substituting values from (1) and (2) in 2e12 + e22


2e12 + e22







= 3


Hence, value of 2e12 + e22 is 3

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