What is the vertex of the parabola $y^2 + 6x - 2y + 13 = 0$?

Three points $O(0,0)$,$P(a, a^2)$,and $Q(-b, b^2)$ with $a > 0$ and $b > 0$ lie on the parabola $y = x^2$. Let $S_1$ be the area of the region bounded by the line $PQ$ and the parabola,and $S_2$ be the area of the triangle $OPQ$. If the minimum value of $\frac{S_1}{S_2}$ is $\frac{m}{n}$,where $\operatorname{gcd}(m, n) = 1$,then $m + n$ is equal to:

The equations $x = \frac{t}{4}$ and $y = \frac{t^2}{4}$ represent:

Find the coordinates of the focus,axis of the parabola,the equation of the directrix,and the length of the latus rectum for $x^{2}=6y$.

What is the focus of the parabola $y^2 - x - 2y + 2 = 0$?

What is the angle in degrees between the tangents drawn from the origin to the parabola $y^2 = 4a(x - a)$?