The point of intersection of the latus rectum and axis of the parabola $y^2+4x+2y-8=0$ is

An equilateral triangle is inscribed in the parabola $y^2 = 8x$,with one of its vertices at the vertex of the parabola. Then,the length of the side of that triangle 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 focus of the parabola $4y^2 - 6x - 4y = 5$ is

The equation of the normals at the ends of the latus rectum of the parabola $y^{2} = 4ax$ is:

Find the measure of the angle in degrees subtended by the double ordinate of the parabola $y^2 = 4ax$ at its vertex.