If the tangent to the curve $y^2 = 4x$ at point $(1, 2)$ cuts the coordinate axes at points $A$ and $B$,then the area of $\Delta AOB$ is (where $'O'$ is the origin).

Statement $-1:$ The line $x - 2y = 2$ meets the parabola $y^2 + 2x = 0$ only at the point $(-2, -2).$
Statement $-2:$ The line $y = mx - \frac{1}{2m}$ $(m \neq 0)$ is tangent to the parabola $y^2 = -2x$ at the point $\left( -\frac{1}{2m^2}, -\frac{1}{m} \right).$

The equation of the given curve is $x^2-4x+4y-8=0$. Match the following:

List-$I$	List-$II$
$(A)$ Focus	$(I)$ $(4,2)$
$(B)$ Vertex	$(II)$ $(3,2)$
$(C)$ One end of the latus rectum	$(III)$ $(2,3)$
$(D)$ Point of intersection of the axis and directrix	$(IV)$ $(2,4)$
	$(V)$ $(2,2)$

The correct matching is:

$A$ normal with slope $\frac{1}{\sqrt{6}}$ is drawn from the point $(0, -\alpha)$ to the parabola $x^2 = -4ay$,where $a > 0$. Let $L$ be the line passing through $(0, -\alpha)$ and parallel to the directrix of the parabola. Suppose that $L$ intersects the parabola at two points $A$ and $B$. Let $r$ denote the length of the latus rectum and $s$ denote the square of the length of the line segment $AB$. If $r : s = 1 : 16$,then the value of $24a$ is. . . .

Find the coordinates of the focus,axis of the parabola,the equation of the directrix,and the length of the latus rectum for $y^{2}=12x$.

The angle of intersection between the curves $x^2 = 4(y + 1)$ and $x^2 = -4(y + 1)$ is