Find the point where the normal at the point $(4, 4)$ intersects the parabola $y^2 = 4x$ again.

If the normal drawn at the point $P(9, 9)$ on the parabola $y^2 = 9x$ meets the parabola again at $Q(a, b)$,then $2a + b =$

The equation of the directrix of the parabola $x^2+8x+12y+4=0$ is

The equation of the directrix of the parabola $y^2+4y+4x+2=0$ is

If $L$ is the normal drawn to the parabola $y^2 = 8x$ at the point $t = \frac{1}{\sqrt{2}}$,then the foot of the perpendicular drawn from the focus of the parabola on to the normal $L$ is

Let $L$ be a normal to the parabola $y^2=4x$. If $L$ passes through the point $(9,6)$,then $L$ is given by
$(A)$ $y-x+3=0$ $(B)$ $y+3x-33=0$ $(C)$ $y+x-15=0$ $(D)$ $y-2x+12=0$