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 it has the least area but contains the circle (x-1)^(2)+y^(2)=1 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 (a) 1-1/(sqrt(2)) (b) 2sqrt(2)-2 (c) sqrt(2)-1 (d) none of these

Find the area of the smaller region bounded by the ellipse (x^2)/(a^2)+(y^2)/(b^2)=1 and the line x/a+y/b=1

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

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) sqrt3/2 (C) 1/sqrt2 (D) 1/sqrt3

The ratio in which the line segment joining points A(a_1,\ b_1) and B(a_2,\ b_2) is divided by y-axis is (a) -a_1: a_2 (b) a_1: a_2 (c) b_1: b_2 (d) -b_1: b_2

If ax^2+2bx+c=0 and a_1x^2+2b_1x+c_1=0 have a common root and a/a_1,b/b_1,c/c_1 are in AP then a_1,b_1,c_1 are in (A) A.P. (B) G.P. (C) H.P. (D) none of these