Find the equation of the tangent to the curve $y^2 = 6x$ at the point $(2, -3)$.

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

From an external point $P(h, k)$,a pair of tangent lines are drawn to the parabola $y^2 = 4x$. If $\theta_1$ and $\theta_2$ are the inclinations of these tangents with the $x$-axis such that $\theta_1 + \theta_2 = \frac{\pi}{4}$,then the locus of $P$ is:

What is the vertex of the parabola $x^2 - 8y - x + 19 = 0$?

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:

If the vertex of a parabola is $(4,3)$ and its directrix is $3x+2y-7=0$,then the equation of the latus rectum of the parabola is