The equation of a curve in polar coordinates is $\frac{l}{r} = 2 \sin^2 \frac{\theta}{2}$. This equation represents:

Let $P$ be a point on the parabola $y^{2}=12x$ and $N$ be the foot of the perpendicular drawn from $P$ on the axis of the parabola. $A$ line is drawn through the mid-point $M$ of $PN$,parallel to the axis of the parabola,which meets the parabola at $Q$. If the $y$-intercept of the line $NQ$ is $\frac{4}{9}$,then:

The nearest point on the curve $x^2=2y$ to the point $(0,5)$ is . . . . . . .

Find the coordinates of the point on the parabola $y^2 = 8x$ whose focal distance is $4$.

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).

Consider the parabola $y^2+2x+2y-3=0$ and match the items of List-$I$ with those of the List-$II$.

$A. \ 2x-5=0$	$I. \ \text{Vertex}$
$B. \ (\frac{3}{2}, -1)$	$II. \ \text{Focus}$
$C. \ y+1=0$	$III. \ \text{Equation of directrix}$
$D. \ (2, -1)$	$IV. \ \text{Equation of the axis}$
	$V. \ \text{Equation of the Latus rectum}$

The correct match is: