The shortest distance from $(0,3)$ to the parabola $y^2=4x$ is

If $mx - y + c = 0$ is a normal at a point $P$ on the parabola $y^2 = 16x$ and the focal distance of $P$ is $40$ units,then $|c| =$

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

If the segment of the parabola $y^2 = 4ax$ intercepted by its directrix subtends an angle $\theta$ at its focus,then $\theta = \dots$

Let $A(0,1)$,$B(1,1)$,and $C(1,0)$ be the mid-points of the sides of a triangle with incentre at the point $D$. If the focus of the parabola $y^2 = 4ax$ passing through $D$ is $(\alpha + \beta \sqrt{2}, 0)$,where $\alpha$ and $\beta$ are rational numbers,then $\frac{\alpha}{\beta^2}$ is equal to

The focus of a parabolic mirror as shown in the figure is at a distance of $5 \, cm$ from its vertex. If the mirror is $45 \, cm$ deep,find the distance $AB$. (in $, cm$)