The parametric equations of the parabola $y^2-8x-4y-12=0$ are

The tangent at the point $(1, 2)$ to the curve $y^2 = 4x$ makes an angle $\theta$ with the positive direction of the $X$-axis. Then $\theta =$ (in $^{\circ}$)

Find the equation of the normal to the parabola $y^2 = 4x$ which is parallel to the line $y = 3x + 4$.

Let $O$ be the vertex and $Q$ be any point on the parabola $x^2=8y$. If the point $P$ divides the line segment $OQ$ internally in the ratio $1:3$, then the locus of $P$ is

If $(2,0)$ is the vertex and the $Y$-axis is the directrix of a parabola,then its focus 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: $P \quad Q \quad R \quad S$