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

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

Show that the tangents at the ends of conjugate diameters of the ellipse x^(2)/a^(2)+y^(2)/b^(2)=1 intersect on the ellipse x^(2)/a^(2)+y^(2)/b^(2)=2 .

If (asectheta, btantheta) and (asecphi, btanphi) be two coordinate of the ends of a focal chord passing through (ae,0) of x^2/a^2-y^2/b^2=1 then tan(theta/2)tan(phi/2) equals to

if 2y = x and 3y + 4x=0 are the equations of a pair of conjugate diameters of an ellipse , then the eccentricity of the ellipse , is

the equation of a diameter conjugate to a diameter y=(b)/(a)x of the ellipse (x^(2))/(a^(2))+(y^(2))/(b^(2))=1 is

The distance of the point 'theta' on the ellipse x^(2)/a^(2) + y^(2)/b^(2) = 1 from a focus, is

If the length of the latusrectum of the ellipse x^(2) tan^(2) theta + y^(2) sec^(2) theta = 1 is 1/2, then theta =

If eccentric angle of a point lying in the first quadrant on the ellipse x^2/a^2 + y^2/b^2 = 1 be theta and the line joining the centre to that point makes an angle phi with the x-axis, then theta - phi will be maximum when theta is equal to

If 2sec 2theta=tan phi+cotphi , then one of the values of theta+phi is