If normal to ellipse `x^2/a^2+y^2/b^2=1` at `(ae,b^2/a)` is passing through `(0,-2b)` then value of eccentricity is

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

The ellipse x^2/a^2+y^2/b^2=1,(a gt b) passes through (2,3) and have eccentricity equal to 1/2 . Then find the equation of normal to the ellipse at (2,3) .

If the focal chord of the ellipse (x^(2))/(a^(2))+(y^(2))/(b^(2))=1, is normal at (a cos theta,b sin theta) then eccentricity of the ellipse is

If equation of directrix of an ellipse x^2/a^2+y^2/b^2=1 is x=4, then normal to the ellipse at point (1,beta),(beta gt 0) passes through the point (where eccentricity of the ellipse is 1/2 )

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.