If two ellipse ` E ( a gt b ) "and " E ( alpha gt beta ) ` have the same eccentricity , then

a = e gt b = d gt c

The tangent at the point alpha on the ellipse x^2/a^2+y^2/b^2=1 meets the auxiliary circle in two points which subtends a right angle at the centre, then the eccentricity 'e' of the ellipse is given by the equation (A) e^2(1+cos^2alpha)=1 (B) e^2(cosec^2alpha-1)=1 (C) e^2(1+sin^2alpha)=1 (D) e^2(1+tan^2alpha)=1

If alpha and beta are the eccentric angles of the extremities of a focal chord of an ellipse, then prove that the eccentricity of the ellipse is (sinalpha+sinbeta)/("sin"(alpha+beta))

If alpha and beta are the eccentric angles of the extremities of a focal chord of an ellipse, then prove that the eccentricity of the ellipse is (sinalpha+sinbeta)/("sin"(alpha+beta))

Area bounded between two latus-rectum of the ellipse (x^(2))/(a^(2))+ (y^(2))/(b^(2))= 1, a gt b is _____ (where, e is eccentricity of the ellipse)

If alpha and beta are the eccentric angles of the extremities of a focal chord of an ellipse,then prove that the eccentricity of the ellipse is (sin alpha+sin beta)/(sin(alpha+beta))

For an ellipse (x^2)/(a^2) +(y^2)/(b^2) =1, a gt b , the area enclosed by two latus rectum is ……. ( where is the eccentricity of an ellipse )