The correct option is(D): \(8\sqrt{3}\).
Let AB = \(\ell , \, then \, \, AM \, = \ell cos 30^{?} \, = \frac{\ell \sqrt{3}}{2}\)
& BM = \(\ell \, sin 30^{?} \, = \frac{\ell}{2}\)

So, the coordinates of B are \(\bigg( \frac{\ell \sqrt{3}}{2}, \frac{\ell}{2}\bigg)\)
Since, B lies on \(y^2 \, = \, 4x\)
\(\therefore \, \, \frac{\ell^2}{4}=4\bigg(\frac{\ell \sqrt{3}}{2}\bigg)\)
\(\Rightarrow \, \, \ell^2 \, \frac{16}{2}. \sqrt{3\ell} \, \, \Rightarrow \, \, =8\sqrt{3}\)
Two parabolas have the same focus $(4, 3)$ and their directrices are the $x$-axis and the $y$-axis, respectively. If these parabolas intersect at the points $A$ and $B$, then $(AB)^2$ is equal to:
Let \( y^2 = 12x \) be the parabola and \( S \) its focus. Let \( PQ \) be a focal chord of the parabola such that \( (SP)(SQ) = \frac{147}{4} \). Let \( C \) be the circle described by taking \( PQ \) as a diameter. If the equation of the circle \( C \) is: \[ 64x^2 + 64y^2 - \alpha x - 64\sqrt{3}y = \beta, \] then \( \beta - \alpha \) is equal to:
Parabola is defined as the locus of points equidistant from a fixed point (called focus) and a fixed-line (called directrix).

=> MP2 = PS2
=> MP2 = PS2
So, (b + y)2 = (y - b)2 + x2