Find the equation of the normal to the ellipse `(x^2)/(a^2)+(y^2)/(b^2)=1` at the positive end of the latus rectum.

An equation of the normal to the ellipse x^2//a^2+y^2//b^2=1 with eccentricity e at the positive end of the latus rectum is

Find the equations of tangent and normal to the ellipse (x^(2))/(a^(2))+(y^(2))/(b^(2))=1 at (x_(1),y_(1))

The value of a for the ellipse (x^(2))/(a^(2))+(y^(2))/(b^(2))=1,(a>b), if the extremities of the latus rectum of the ellipse having positive ordinates lie on the parabola x^(2)=2(y-2) is

The normal drawn to the ellipse (x^(2))/(a^(2))+(y^(2))/(b^(2))=1 at the extremity of the latus rectum passes through the extremity of the minor axis.Eccentricity of this ellipse is equal

For the ellipse (x^(2))/(4) + (y^(2))/(3) = 1 , the ends of the two latus rectum are the four points

Find the length of the latus rectum of the ellipse x^2/36+y^2/25=1

Find the length of the latus -rectum of the ellipse : (x^(2))/(4) + (y^(2))/(9) = 1 .