Find the vertex of the parabola $y^2 + 6x - 2y + 13 = 0$.

One of the points on the parabola $y^2 = 12x$ with focal distance $12$ is:

If the normal chord drawn at the point $\left(\frac{15}{2}, \frac{15}{\sqrt{2}}\right)$ to the parabola $y^2=15x$ subtends an angle $\theta$ at the vertex of the parabola,then $\sin \frac{\theta}{3}+\cos \frac{2\theta}{3}-\sec \frac{4\theta}{3}=$

The line $y - \sqrt{3}x + 3 = 0$ cuts the parabola $y^2 = x + 2$ at the points $P$ and $Q$. If the coordinates of the point $X$ are $(\sqrt{3}, 0)$, then the value of $XP \cdot XQ$ is

For the parabola $y^2+6y-2x=-5$,consider the following statements:
$I$. The vertex is $(-2, -3)$.
$II$. The directrix is $y+3=0$.
Which of the following is correct?

If a normal chord at a point $t (\neq 0)$ on the parabola $y^2 = 9x$ subtends a right angle at its vertex,then $t =$