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 one end of a latus rectum of the ellipse x^2/a^2+y^2/b^2=1 passes through one end of the minor axis, then show that e^4+e^2=1 [ e is the eccentricity of the ellipse]

If the normal at one end of latusrectum of an ellipse (x^(2))/(a^(2))+(y^(2))/(b^(2))=1 passes through one end of minor axis then

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.

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.

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