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

The area of the ellipse (x^(2))/(a^(2))+(y^(2))/(b^(2))=1 is

The ellipse (x^(2))/(a^(2))+(y^(2))/(b^(2))=1 is such that its has the least area but contains the circel (x-1)^(2)+y^(2)=1 The eccentricity of the ellipse is

If the area of the auxiliary circle of the ellipse (x ^(2))/(a ^(2)) + (y ^(2))/(b ^(2)) =1 (a gt b) is twice the area of the ellipse, then the eccentricity of the ellipse is

A parabola is drawn with focus at one of the foci of the ellipse (x^(2))/(a^(2))+(y^(2))/(b^(2))=1. If the latus rectum of the ellipse and that of the parabola are same,then the eccentricity of the ellipse is 1-(1)/(sqrt(2)) (b) 2sqrt(2)-2sqrt(2)-1( d) none of these

Pa n dQ are the foci of the ellipse (x^2)/(a^2)+(y^2)/(b^2)=1 and B is an end of the minor axis. If P B Q is an equilateral triangle, then the eccentricity of the ellipse is 1/(sqrt(2)) (b) 1/3 (d) 1/2 (d) (sqrt(3))/2

Prove that the area bounded by the circle x^(2)+y^(2)=a^(2) and the ellipse (x^(2))/(a^(2))+(y^(2))/(b^(2))=1 is equal to the area of another ellipse having semi-axis a-b and b,a>b

The ratio of the area of triangle inscribed in ellipse (x^(2))/(a^(2))+(y^(2))/(b^(2))=1 to that of triangle formed by the corresponding points on the auxiliary circle is 0.5. Then,find the eccentricity of the ellipse.(A) (1)/(2) (B) (sqrt(3))/(2) (C) (1)/(sqrt(2))(D)(1)/(sqrt(3))