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

The shortest distance between the line $y-x=1$ and the curve $x=y^2$ is

Let $P$ and $Q$ be points on the parabola $y^{2}=4x$ such that the line segment $PQ$ subtends a right angle at the vertex. If $PQ$ intersects the axis of the parabola at $R$, then the distance of the vertex from $R$ 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

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

Let $A(4, -4)$ and $B(9, 6)$ be points on the parabola $y^2 = 4x$. Let $C$ be a point chosen on the arc $AOB$ of the parabola,where $O$ is the origin,such that the area of $\Delta ACB$ is maximum. Then,the area (in sq. units) of $\Delta ACB$ is