The eccentricity of the parabola $x^2 - 4x - 4y + 4 = 0$ is

Let the parabola $y = x^2 + px + q$ passing through the point $(1, -1)$ be such that the distance between its vertex and the $x$-axis is minimum. Then the value of $p^2 + q^2$ is:

If the normal to the parabola $y^2=4x$ at $P(1,2)$ meets the parabola again at $Q$,then the coordinates of $Q$ are

The point of intersection of the tangents to the parabola $y^2 = 4ax$ at the points $t_1$ and $t_2$ is

The point $(3,4)$ is the focus and $2x - 3y + 5 = 0$ is the directrix of a parabola. Its latus rectum is:

Find the equation of the normal to the parabola $y^2 = 16x$ having a slope of $-1/4$.