The point of intersection of the tangents to the parabola $y^2 = 4ax$ at the points $t_1$ and $t_2$ is

The position of a moving point in the $XY$-plane at time $t$ is given by $\left( (u \cos \alpha)t, (u \sin \alpha)t - \frac{1}{2}gt^2 \right)$,where $u, \alpha, g$ are constants. The locus of the moving point is

Let $P : y^{2} = 4ax, a > 0$ be a parabola with focus $S$. Let the tangents to the parabola $P$ that make an angle of $\frac{\pi}{4}$ with the line $y = 3x + 5$ touch the parabola $P$ at $A$ and $B$. Then the value of $a$ for which $A, B$ and $S$ are collinear is:

If a normal chord at a point $t (\neq 0)$ on the parabola $y^2 = 9x$ subtends a right angle at its vertex,then $t =$

$A$ 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 $A$ be the vertex and $L$ be the length of the latus rectum of the parabola $y^2 - 2y - 4x - 7 = 0$. Find the equation of the parabola with $A$ as the vertex,$2L$ as the length of the latus rectum,and the axis at right angles to that of the given curve.