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

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

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 minor axis, then e^(4)+e^(2)= ( where e is the eccentricity of the ellipse)

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.

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.