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 normal at an end of a latus rectum of the ellipse (x^(2))/(a^(2))+(y^(2))/(b^(2))=1 passes through an end of the minor axis if

The normal at an end of a latus rectum of the ellipse (x^(2))/(a^(2))+(y^(2))/(b^(2))=1 passes through an end of the minor axis if

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

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)

If the normal at an end of latus rectum of the hyperbola x^(2)/a^(2) - y^(2)/b^(2) = 1 passes through the point (0, 2b) then

If the normal at the end of latus rectum of the ellipse (x^(2))/(a^(2))+(y^(2))/(b^(2))=1 passes through (0,b) then e^(4)+e^(2) (where e is eccentricity) equals

If the normals at an end of a latus rectum of an ellipse passes through the other end of the minor axis, then prove that e^(4) + e^(2) =1.