The locus of a point which divides the line segment joining the point $(0,-1)$ and a point on the parabola $x^{2}=4y$ internally in the ratio $1:2$ 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 point at which the line $y = mx + c$ touches the parabola $y^2 = 4ax$ is

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If the normal drawn at $P(8, 16)$ to the parabola $y^2 = 32x$ meets the parabola again at $Q$,then the equation of the tangent drawn at $Q$ to the parabola is

What is the focus of the parabola $4y^2 + 12x - 20y + 67 = 0$?