What is the length of the latus rectum of the parabola $x^2 - 4x - 8y + 12 = 0$?

The length of the chord of the parabola $y^{2}=4ax$ $(a>0)$ which passes through the vertex and makes an acute angle $\alpha$ with the axis of the parabola is

The locus of the point of intersection of the perpendicular tangents to the curve $y^2 + 4y - 6x - 2 = 0$ is:

Let $L_1, L_2$ be the lines passing through the point $P(0,1)$ and touching the parabola $9x^2+12x+18y-14=0$. Let $Q$ and $R$ be the points on the lines $L_1$ and $L_2$ such that the $\triangle PQR$ is an isosceles triangle with base $QR$. If the slopes of the lines $QR$ are $m_1$ and $m_2$,then $16(m_1^2+m_2^2)$ is equal to ..............

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

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: