Concyclic points on ellipse

If alpha,beta,gamma,delta be eccentric angles ofthe four concyclic points ofthe ellipse (x^(2))/(a^(2))+(y^(2))/(b^(2))=1 then alpha+beta+gamma+delta=(A)(2n+1)(pi)/(2) (B) (2n-1)pi(C)2n pi(D)n pi

Concyclic points on hyperbola

Q is a point on the auxiliary circle of an ellipse. P is the corresponding point on ellipse. N is the foot of perpendicular from focus S, to the tangent of auxiliary circle at Q. Then

In the figure , points G,D,E,F are concyclic points of a cicle with centreC. /_ECF = 70^(@) m ( arc DGF ) = 200^(@) , find m(arc DE) and m(arc DEF ) .

In the adjoining figure A, B, C, D are the concyclic points. The value of 'x' is :

Derive an equation to find distance of a point on ellipse to the focus of ellipse.

Number of points on the ellipse x^2/a^2+y^2/b^2=1 at which the normal to the ellipse passes through at least one of the foci of the ellipse is