`P` and `Q` 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

S and T 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 STB is an equilateral triangle, 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

P is a variable on the ellipse (x^2)/(a^2)+(y^2)/(b^2)=1 with AA' as the major axis. Find the maximum area of triangle A P A '

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

Chords of the ellipse (x^2)/(a^2)+(y^2)/(b^2)=1 are drawn through the positive end of the minor axis. Then prove that their midpoints lie on the ellipse.

S and T are the foci of an ellipse and B is an end point of the minor axis . IF /_\STB is equilateral then e =

If the length of the major axis intercepted between the tangent and normal at a point P (a cos theta, b sin theta) on the ellipse (x^(2))/(a^(2)) +(y^(2))/(b^(2)) =1 is equal to the length of semi-major axis, then eccentricity of the ellipse is

An ellipse has OB, as semi minor axis, F and F' its foci and the angle FBF' is a right angle. Then the eccentricity of the ellipse is :

Extremities of the latera recta of the ellipses (x^2)/(a^2)+(y^2)/(b^2)=1(a > b) having a given major axis 2a lies on

If the normal at one end of the latus rectum of the ellipse (x^2)/(a^2)+(y^2)/(b^2)=1 passes through one end of the minor axis, then prove that eccentricity is constant.