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`.

The area of the ellipse (x^(2))/(a^(2))+(y^(2))/(b^(2))=1 is

The locus of the poles of tangents to the director circle of the ellipse x^(2)/a^(2) + y^(2)/b^(2) = 1 with respect to the ellipse x^(2)/a^(2) + y^(2)/b^(2) = 1 is

If CP and CD are semi-conjugate diameters of the ellipse x^(2)/a^(2) + y^(2)/b^(2) = 1 , then CP^(2) + CD^(2) =

The line x = at^(2) meets the ellipse x^(2)/a^(2) + y^(2)/b^(2) = 1 in the real points iff

Two perpendicular tangents drawn to the ellipse (x^2)/(25)+(y^2)/(16)=1 intersect on the curve.

CP and CD are conjugate semi-diameters of the ellipse x^(2)/a^(2) + y^(2)/b^(2) = 1 , The locus of the mid-point of PD, is

For the ellipse (x^(2))/(a^(2))+(y^(2))/(b^(2)) =1 and (x^(2))/(b^(2))+(y^(2))/(a^(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 e is eccentricity of the ellipse (x^(2))/(a^(2))+(y^(2))/(b^(2))=1 (where,a lt b), then

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