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

Let $$a$$ and $$b$$ be non-zero real numbers. Then, the equation $${x^2}{\cos ^2}\theta - {y^2}{\sin ^2}\theta = 0$$ represents

STATEMENT-1: The curve $$y = {{ - {x^2}} \over 2} + x + 1$$ is symmetric with respect to the line $$x=1$$. because

STATEMENT-2: A parabola is symmetric about its axis.

Consider the circle $${x^2} + {y^2} = 9$$ and the parabola $${y^2} = 8x$$. They intersect at $$P$$ and $$Q$$ in the first and the fourth quadrants, respectively. Tangent to the circle at $$P$$ and $$Q$$ intersect the $$x$$-axis at $$R$$ and tangents to the parabola at $$P$$ and $$Q$$ intersect the $$x$$-axis at $$S$$.

The radius of the incircle of the triangle $$PQR$$ is