Let $$y^{2} = 12x$$ be the parabola and S be its focus. Let PQ be a focal chord of the parabola such that (SP)(SQ) = $$\frac{147}{4}$$. Let C be the circle described taking PQ as a diameter. If the equation of a circle C is $$64x^2 + 64y^2 - \alpha x - 64\sqrt{3}\,y = \beta$$, then $$\beta - \alpha$$ is equal to ________.

The given parabola is $$y^{2}=12x$$. Write it as $$y^{2}=4ax$$ to identify $$a$$.

$$4a=12 \;\;\Longrightarrow\;\; a=3$$

For the parabola $$y^{2}=4ax$$: • Focus $$S(a,0)$$. • A point on the parabola in parametric form is $$P(t):\; (at^{2},\,2at)$$. Thus here a general point is $$\bigl(3t^{2},\,6t\bigr)$$.

If $$P(t_{1})$$ and $$Q(t_{2})$$ are the end-points of a focal chord, then $$t_{1}t_{2}=-1$$. (Standard result for a focal chord of $$y^{2}=4ax$$.)

Coordinates of the required points: $$P\;(3t_{1}^{2},\,6t_{1}), \qquad Q\;(3t_{2}^{2},\,6t_{2})$$

Distance of $$P$$ from the focus $$S(3,0)$$:

$$\begin{aligned} SP^{2}&=(3t_{1}^{2}-3)^{2}+(6t_{1})^{2} \\ &=9\bigl[(t_{1}^{2}-1)^{2}+(2t_{1})^{2}\bigr] \\ &=9(t_{1}^{4}+2t_{1}^{2}+1)=9(t_{1}^{2}+1)^{2} \end{aligned}$$

So $$SP = 3\,(t_{1}^{2}+1)$$. Similarly, $$SQ = 3\,(t_{2}^{2}+1)$$.

The problem states $$SP\cdot SQ=\dfrac{147}{4}$$, hence

$$9\,(t_{1}^{2}+1)(t_{2}^{2}+1)=\dfrac{147}{4} \;\;\Longrightarrow\;\; (t_{1}^{2}+1)(t_{2}^{2}+1)=\dfrac{49}{12}\; -(1)$$

Using $$t_{1}t_{2}=-1$$, put $$t_{2}=-\dfrac{1}{t_{1}}$$ in $$(1)$$. Let $$x=t_{1}^{2}\,(\gt 0)$$. Then

$$\frac{(x+1)^{2}}{x}=\frac{49}{12} \;\;\Longrightarrow\;\;12(x+1)^{2}=49x$$

$$\Rightarrow 12x^{2}-25x+12=0 \;\;\Longrightarrow\;\; x=\frac{32}{24}=\frac{4}{3}\quad\text{or}\quad x=\frac{18}{24}=\frac{3}{4}$$

Choosing $$t_{1}^{2}=\dfrac{4}{3}$$ (the other root only interchanges $$P$$ and $$Q$$):

$$t_{1}= \frac{2}{\sqrt3},\qquad t_{2}=-\frac{\sqrt3}{2}$$

Coordinates of the chord end-points:

$$\begin{aligned} P &: (3t_{1}^{2},\,6t_{1})=(4,\;4\sqrt3)\\ Q &: (3t_{2}^{2},\,6t_{2})=(\tfrac94,\;-3\sqrt3) \end{aligned}$$

Centre and radius of circle with $$PQ$$ as diameter

Mid-point $$M\bigl(\dfrac{4+\tfrac94}{2},\;\dfrac{4\sqrt3-3\sqrt3}{2}\bigr) = \Bigl(\dfrac{25}{8},\;\dfrac{\sqrt3}{2}\Bigr)$$

Length $$PQ$$:

$$\begin{aligned} PQ^{2}&=\Bigl(4-\tfrac94\Bigr)^{2} + \bigl(4\sqrt3+3\sqrt3\bigr)^{2}\\ &=\Bigl(\tfrac74\Bigr)^{2} + (7\sqrt3)^{2} = \frac{49}{16}+147=\frac{2401}{16}\\ \Rightarrow PQ&=\frac{49}{4} \end{aligned}$$

Radius $$r=\dfrac{PQ}{2}=\dfrac{49}{8},\qquad r^{2}=\Bigl(\dfrac{49}{8}\Bigr)^{2}=\dfrac{2401}{64}$$

Equation of the circle

Using centre-radius form: $$(x-\tfrac{25}{8})^{2}+(y-\tfrac{\sqrt3}{2})^{2}=r^{2}$$

Expanding & regrouping as $$x^{2}+y^{2}+Dx+Ey+F=0$$ gives

$$x^{2}+y^{2}-\frac{25}{4}\,x-\sqrt3\,y+\Bigl(\frac{625}{64}+\frac{3}{4}-\frac{2401}{64}\Bigr)=0$$

The constant term evaluates to $$F=-27$$, hence

$$x^{2}+y^{2}-\frac{25}{4}\,x-\sqrt3\,y-27=0$$

Multiply by 64 to match the given pattern:

$$64x^{2}+64y^{2}-400x-64\sqrt3\,y=1728$$

Comparing with $$64x^{2}+64y^{2}-\alpha x-64\sqrt3\,y=\beta$$, we get $$\alpha=400,\qquad\beta=1728$$

Therefore, $$\beta-\alpha = 1728-400 = 1328$$.

Final Answer : 1328