The coordinates of a moving point $P$ are $(2t^2+4, 4t+6)$. Then its locus will be a

Find the equation of the parabola which passes through $(6,-2)$,has its vertex at the origin and its axis along the $y$-axis.

$A$ quadratic polynomial $y = f(x)$ with absolute term $3$ neither touches nor intersects the abscissa axis and is symmetric about the line $x = 1$. The coefficient of the leading term of the polynomial is unity. $A$ point $A(x_1, y_1)$ with abscissa $x_1 = 1$ and a point $B(x_2, y_2)$ with ordinate $y_2 = 11$ are given in a Cartesian rectangular system of coordinates $OXY$ in the first quadrant on the curve $y = f(x)$,where $O$ is the origin. The graph of $y = f(x)$ represents a parabola whose focus has the coordinates:

From a point $(d, 0)$, three normals are drawn to the parabola $y^{2} = x$. Then:

If the line $2bx + 3cy + 4d = 0$ passes through the points of intersection of $y^2 = 4ax$ and $x^2 = 4ay$,then

If two parabolas $y^2 = 4x$ and $x^2 = 32y$ intersect at the point $(16, 8)$ at an angle $\theta$,then what is the value of $\tan \theta$?