Let y^2=12x have focus S, and let PQ be a focal chord such that SP SQ=1474.

Question

Let $y^2=12x$ have focus $S$, and let $PQ$ be a focal chord such that $SP\cdot SQ=\dfrac{147}{4}$. Let $C$ be the circle described taking $PQ$ as a diameter. If the equation of $C$ is $64x^2+64y^2-\alpha x-64\sqrt3 y=\beta$, then $\beta-\alpha$ is equal to \_\_\_\_\_\_\_\_.

Answer: 1328

Accepted Answer

Here $a=3$; $P=(3t^2,6t),\ Q=\left(\dfrac{3}{t^2},-\dfrac{6}{t}\right)$ (using $t_1t_2=-1$).

$$SP\cdot SQ = 9\left(t^2+\frac{1}{t^2}+2\right)=\frac{147}{4} \;\Rightarrow\; t^2+\frac{1}{t^2}=\frac{25}{12}.$$

Circle on diameter $PQ$: $x^2+y^2-3\left(t^2+\dfrac1{t^2}\right)x-6\left(t-\dfrac1t\right)y+(x_Px_Q+y_Py_Q)=0$, with $x_Px_Q=9,\ y_Py_Q=-36$:

$$x^2+y^2-3\left(t^2+\frac1{t^2}\right)x-6\left(t-\frac1t\right)y-27=0.$$

Multiplying by $64$: $\alpha=192\left(t^2+\dfrac1{t^2}\right)=192\times\dfrac{25}{12}=400$, and the constant term gives $\beta=1728$.

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

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