The normal drawn to the ellipse `x^2/a^2+y^2/b^2=1` at the extremity of the latus rectum passes through the extremity of the minor axis. Eccentricity of this ellipse is equal to

If the normal at an end of a latus rectaum of an ellipse passes through an extremity of the minor axis then the eccentricity of the ellispe satisfies .

If latus-rectum is one-third minor axis, then eccentricity of the ellipse is

If the normal at an end of a latus-rectum of an ellipse x^(2)/a^(2) + y^(2)/b^(2) = 1 passes through one extremity of the minor axis, show that the eccentricity of the ellipse is given by e = sqrt((sqrt5-1)/2)

The latus rectum of an ellipse is half of its minor axis. Its eccentricity is :

If the normal at an end of a latus-rectum of an elipse (x^(2))/(a^(2))+(y^(2))/(b^(2))=1 passes through one extremity of the minor axis,show that the eccentricity of the ellipse is given by e^(4)+e^(-1)=0

The normla to the hyperbola (x^2)/(a^2)-(y^2)/(b^2)=1 drawn at an extremity of its latus rectum is parallel to an asymptote. Show that the eccentricity is equal to the square root of (1+sqrt(5))//2.

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