The area of triangle formed by the lines joining the vertex of the parabola, x^2=8 y, to the extremities of its latus rectum is

Suppose that two chords, drawn from the point $(1, 2)$ on the circle $x^2 + y^2 + x - 3y = 0$ are bisected by the $y$-axis. If the other ends of these chords are $R$ and $S$, and the mid point of the line segment $RS$ is $(\alpha, \beta)$, then $6(\alpha + \beta)$ is equal to:

Let $\dfrac{x^2}{f(a^2+7a+3)} + \dfrac{y^2}{f(3a+15)} = 1$ represent an ellipse with major axis along $y$-axis, where $f$ is a strictly decreasing positive function on $\mathbb{R}$. If the set of all possible values of $a$ is $\mathbb{R} - [\alpha, \beta]$, then $\alpha^2+\beta^2$ is equal to:

Let the vertex $A$ of a triangle $ABC$ be $(1, 2)$, and the mid-point of the side $AB$ be $(5, -1)$. If the centroid of this triangle is $(3, 4)$ and its circumcenter is $(\alpha, \beta)$, then $21(\alpha + \beta)$ is equal to:

Let O be the origin, and P and Q be two points on the rectangular hyperbola $xy = 12$ such that the mid point of the line segment PQ is $\left(\dfrac{1}{2}, -\dfrac{1}{2}\right)$. Then the area of the triangle OPQ equals:

If the line $\alpha x+4 y=\sqrt{7}$, where $\alpha \in \mathbf{R}$, touches the ellipse $3 x^{2}+4 y^{2}=1$ at the point P in the first quadrant, then one of the focal distances of $P$ is :