If the normal at one end of lotus rectum of an ellipse passes through one end of minor axis then prove that,
`e^(4)+e^(2)-1=0`

If the normal at one end of lotus rectum of an ellipse passes through one end of minor axis then prove that, e^(2)=(sqrt(5)-1)/(2) , where e is the eccentricity of the ellipse.

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.

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.

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 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 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 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 latus rectum of an ellipse is half of its minor axis then e =

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.