If $x-y-3=0$ is a normal drawn through the point $(5,2)$ to the parabola $y^2=4x$,then the slope of the other normal that can be drawn through the same point to the parabola $y^2=4x$ is

$A$ line $L: y=mx+3$ meets the $y$-axis at $E(0,3)$ and the arc of the parabola $y^2=16x, 0 \leq y \leq 6$ at the point $F(x_0, y_0)$. The tangent to the parabola at $F(x_0, y_0)$ intersects the $y$-axis at $G(0, y_1)$. The slope $m$ of the line $L$ is chosen such that the area of the triangle $EFG$ has a local maximum.
Match List $I$ with List $II$ and select the correct answer using the code given below the lists:

List $I$	List $II$
$P. \quad m=$	$1. \quad 1/2$
$Q. \quad \text{Maximum area of } \triangle EFG \text{ is}$	$2. \quad 4$
$R. \quad y_0=$	$3. \quad 2$
$S. \quad y_1=$	$4. \quad 1$

Codes: $P \quad Q \quad R \quad S$

The number of points on the parabola $y^2=x$ at which the slope of the normal drawn at the point is equal to the $x$-coordinate of that point is

If the straight line $y=mx+c$ is parallel to the axis of the parabola $y^2=lx$ and intersects the parabola at $\left(\frac{c^2}{8}, c\right)$,then the length of the latus rectum is

If one end of a focal chord $AB$ of the parabola $y^{2}=8x$ is at $A\left(\frac{1}{2},-2\right)$,then the equation of the tangent to it at $B$ is:

The length of the chord of the parabola $y^2 = x$ which is bisected at the point $(2, 1)$ is