Let $y^2 = 12x$ be the parabola and $S$ be its focus. Let $PQ$ be a focal chord of the parabola such that $(SP)(SQ) = \tfrac{147}{4}$. 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 _____.
Step-by-Step Solution
Key Concept: Use $SP\cdot SQ = a^2(1+t^2)(1+1/t^2)$ to find the parameter $t$; write the diametric circle through $P$ and $Q$; clear fractions and match coefficients against the given form.
$y^2=12x$: $a=3$, $S=(3,0)$. Let $P=(3t^2,6t)$; focal chord gives $Q=(3/t^2,-6/t)$. $SP\cdot SQ=9(1+t^2)(1+1/t^2)=\tfrac{147}{4}$, so $\tfrac{(1+t^2)^2}{t^2}=\tfrac{49}{12}$, giving $12t^4-25t^2+12=0$, whence $t^2=\tfrac{3}{4}$ or $\tfrac{4}{3}$. Taking $t=-\tfrac{\sqrt{3}}{2}$: $P=\left(\tfrac{9}{4},-3\sqrt{3}\right)$, $Q=(4,4\sqrt{3})$. Diametric circle: $\left(x-\tfrac{9}{4}\right)(x-4)+(y+3\sqrt{3})(y-4\sqrt{3})=0$, i.e., $x^2+y^2-\tfrac{25}{4}x-\sqrt{3}\,y-27=0$. Multiplying by $64$: $64x^2+64y^2-400x-64\sqrt{3}\,y=1728$. So $\alpha=400$, $\beta=1728$, $\beta-\alpha=1328$.
Correct Answer: 1328