The equation of the normal drawn to the parabola $y^2=6x$ at the point $(24,12)$ is

The axis of the parabola $x^2 - 4x - 3y + 10 = 0$ 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
$(A)$ $y-x+3=0$ $(B)$ $y+3x-33=0$ $(C)$ $y+x-15=0$ $(D)$ $y-2x+12=0$

Let $P, Q,$ and $R$ be three co-normal points on the parabola $y^2 = 4ax$. Then the correct statement$(s)$ is/are:

The normal drawn at a point $(2, -4)$ on the parabola $y^2 = 8x$ cuts the same parabola again at $(\alpha, \beta)$. Then $\alpha + \beta =$

$S \equiv y^2 - 4ax = 0$ and $S' \equiv y^2 + ax = 0$ are two parabolas,and $P(t)$ is a point on the parabola $S' = 0$. If $A$ and $B$ are the feet of the perpendiculars from $P$ onto the coordinate axes and $AB$ is a tangent to the parabola $S = 0$ at the point $Q(t_1)$,then $t_1 =$