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Given a > b >0,a1> b1>0, area of ellipse...

Given `a > b >0,a_1> b_1>0,` area of ellipse `(x^2)/(a^2)+(y^2)/(b^2)=1` is twice the area of ellipse `(x^2)/(a_1^ 2)+(y^2)/(b_1^ 2)=1` . If eccentricity of the ellipse is same then:
(A) `a=sqrt(2)a_1`
(B) `a a_1=bb-1`
(C) `a b=a_1b_1`
(D) none of these

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