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 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

If the eccentricity of the ellipse, x^2/(a^2+1)+y^2/(a^2+2)=1 is 1/sqrt6 then latus rectum of ellipse is

Find the equation of the normal to the ellipse (x^2)/(a^2)+(y^2)/(b^2)=1 at the positive end of the latus rectum.

Find the length of Latus rectum of the ellipse (x^(2))/(a^(2))+(y^(2))/(b^(2))=1 .

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 . . .

Find the eccentricity of an ellipse (x^2)/(a^2)+(y^2)/(b^2)=1 whose latus rectum is half of its major axis. (agtb)

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.

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.

Let Sa n dS ' be two foci of the ellipse (x^2)/(a^2)+(y^2)/(b^2)=1 . If a circle described on S S^(prime) as diameter intersects the ellipse at real and distinct points, then the eccentricity e of the ellipse satisfies (a) c=1/(sqrt(2)) (b) e in (1/(sqrt(2)),1) (c) e in (0,1/(sqrt(2))) (d) none of these