The tangents drawn at the extremities of a focal chord of the parabola $y^{2}=16x$:

What is the vertex of the parabola $9x^2 - 6x + 36y + 9 = 0$?

The equation of the parabola with its vertex at $(1, 1)$ and focus at $(3, 1)$ is

Let $A(0,1)$,$B(1,1)$,and $C(1,0)$ be the mid-points of the sides of a triangle with incentre at the point $D$. If the focus of the parabola $y^2 = 4ax$ passing through $D$ is $(\alpha + \beta \sqrt{2}, 0)$,where $\alpha$ and $\beta$ are rational numbers,then $\frac{\alpha}{\beta^2}$ is equal to

Let $S$ be the set of all $a \in \mathbb{N}$ such that the area of the triangle formed by the tangent at the point $P(b, c)$,where $b, c \in \mathbb{N}$,on the parabola $y^2 = 2ax$ and the lines $x = b$ and $y = 0$ is $16 \text{ unit}^2$. Then $\sum_{a \in S} a$ is equal to $..........$.

If the focal chord drawn through the point $(1,2)$ to the parabola $y^2=8x$ meets this parabola in $(x_1, y_1)$ and $(x_2, y_2)$,then $x_1+x_2=$