Consider a branch of the hyperbola $$${x^2} - 2{y^2} - 2\sqrt 2 x - 4\sqrt 2 y - 6 = 0$$$

with vertex at the point $$A$$. Let $$B$$ be one of the end points of its latus rectum. If $$C$$ is the focus of the hyperbola nearest to the point $$A$$, then the area of the triangle $$ABC$$ is

A hyperbola, having the transverse axis of length $$2\sin \theta ,$$ is confocal with the ellipse $$3{x^2} + 4{y^2} = 12.$$ Then its equation is

If the line $$62x + \sqrt 6 y = 2$$ touches the hyperbola $${x^2} - 2{y^2} = 4$$, then the point of contact is

For hyperbola $${{{x^2}} \over {{{\cos }^2}\alpha }} - {{{y^2}} \over {{{\sin }^2}\alpha }} = 1$$ which of the following remains constant with change in $$'\alpha '$$