Find the eccentric angles of the extremities of the latus recta of the ellipse `(x^2)/(a^2)+(y^2)/(b^2)=1`

The eccentric angles of the ends of L.R. of the ellipse (x^(2)/(a^(2))) + (y^(2)/(b^(2))) = 1 is

If theta and phi are eccentric angles of the ends of a pair of conjugate diametres of the ellipse x^2/a^2 + y^2/b^2 = 1 , then (theta - phi) is equal to

The locus of extremities of the latus rectum of the family of ellipse b^2x^2+a^2y^2=a^2b^2 is

The locus of extremities of the latus rectum of the family of ellipse b^2x^2+a^2y^2=a^2b^2 is

The eccentricity of the ellipse (x^(2))/(36)+(y^(2))/(25)=1 is

The eccentricity of the ellipse (x^(2))/(49)+(y^(2))/(25)=1 is

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

The eccentricity of the ellipse (x^(2))/(16)+(y^(2))/(25) =1 is

Find the eccentricity, foci and the length of the latusrectum of the ellipse x^2+4y^2+8y-2x+1=0.

Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse. 16 x^2+y^2=16