The Cartesian coordinates of the point on the parabola $y^2 = -16x$,whose parameter is $t = \frac{1}{2}$,are

At which point does the line $x + y = 1$ intersect the parabola $y = x - x^2$?

If the normal chord drawn at $(2 a, 2 a \sqrt{2})$ on the parabola $y^2=4 a x$ subtends an angle $\theta$ at its vertex,then $\theta=$ (in $^{\circ}$)

If $a > 0$ and $b^2 - 4ac = 0$,then the curve $y = ax^2 + bx + c$

If two tangents drawn from a point $P$ to the parabola $y^{2}=16(x-3)$ are at right angles,then the locus of point $P$ is :

Let the line $y - \sqrt{3}x + 3 = 0$ cut the parabola $2y^2 = 2x + 3$ at $A$ and $B$. If $P(\sqrt{3}, 0)$,then the value of $|PA - PB|$ is [where $PA$ denotes the distance between points $P$ and $A$].