Let be the parabola and S be its focus. Let PQ be a focal chord of the parabola such that (SP)(SQ) = . Let C be the circle described taking PQ as a diameter. If the equation of a circle C is , then is equal to __________. [2025]

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Solution

Q.
Let $y^{2} = 12 x$ 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 $64 x^{2} + 64 y^{2} - α x - 64 \sqrt{3} y = β$ , then $β - α$ is equal to __________. [2025]

Ans.
(1328)

Given, equation of parabola is $y^{2} = 12 x$

Focus = S = (3, 0)

Let $P (3 t_{1}^{2}, 6 t_{1})$ and $Q (3 t_{2}^{2}, 6 t_{2})$ are points on parabola

Also, $t_{1} t_{2} = - 1$

$∴ P \equiv (3 t^{2}, 6 t) and Q \equiv (\frac{3}{t^{2}}, \frac{- 6}{t})$

Now, $(S P) (S Q) = \frac{147}{4} \Rightarrow (3 + 3 t^{2}) (3 + \frac{3}{t^{2}}) = \frac{147}{4}$

( $∵$ (SP) = PM and (SQ) = QN, where PM and QN are perpendicular distance from directrix)

$\Rightarrow \frac{{(1 + t^{2})}^{2}}{t^{2}} = \frac{49}{12} \Rightarrow 12 t^{4} - 25 t^{2} + 12 = 0$

$\Rightarrow t^{2} = \frac{3}{4}, \frac{4}{3} \Rightarrow t = \pm \frac{\sqrt{3}}{2} or t = \pm \frac{2}{\sqrt{3}}$

Case I : When $t = \frac{\sqrt{3}}{2} or t = - \frac{2}{\sqrt{3}}$

Points are $P (\frac{9}{4}, 3 \sqrt{3}) and Q (4, - 4 \sqrt{3})$

$∴$ Equation of circle is

$(x - 4) (x - \frac{9}{4}) + (y - 3 \sqrt{3}) (y + 4 \sqrt{3}) = 0$

$\Rightarrow x^{2} + y^{2} - \frac{25 x}{4} + \sqrt{3} y - 27 = 0$

On comparing with given equation of circle $64 x^{2} + 64 y^{2} - α x$

$- 64 \sqrt{3} y = β$ , we get $α$ = 400, $β$ = 1728

Case II : When $t = \frac{- \sqrt{3}}{2} or t = \frac{2}{\sqrt{3}}$

Point are $P (\frac{9}{4}, - 3 \sqrt{3}) and Q (4, 4 \sqrt{3})$

Similarly, we get $α$ = 400, $β$ = 1728

$∴ β - α$ = 1728 – 400 = 1328.

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