Let y^2=12x be a parabola and S be its focus. | vedclass

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(A) For the parabola $y^2=12x$,we have $4a=12$,so $a=3$. The focus $S$ is $(3,0)$.Let the coordinates of $P$ be $(3t^2, 6t)$ and $Q$ be $(3/t^2, -6/t)

Vedclass · Accepted Answer

$1328$

Vedclass · Answer

(A) For the parabola $y^2=12x$,we have $4a=12$,so $a=3$. The focus $S$ is $(3,0)$.Let the coordinates of $P$ be $(3t^2, 6t)$ and $Q$ be $(3/t^2, -6/t)$ where $t_1 t_2 = -1$.The distance of a point $(3t^2, 6t)$ from the directrix $x=-3$ is $SP = 3t^2+3$.Similarly,$SQ = 3/t^2+3$.Given $(SP)(SQ) = (3t^2+3)(3/t^2+3) = 9(t^2+1)(1/t^2+1) = 9\frac{(t^2+1)^2}{t^2} = \frac{147}{4}$.$\frac{(t^2+1)^2}{t^2} = \frac{147}{36} = \frac{49}{12}$.Solving for $t^2$,we get $t^2 = 3/4$ or $t^2 = 4/3$.Taking $t^2 = 3/4$,we have $P = (9/4, 3\sqrt{3})$ and $Q = (4, -4\sqrt{3})$.The equation of the circle with diameter $PQ$ is $(x-x_1)(x-x_2) + (y-y_1)(y-y_2) = 0$.$(x-9/4)(x-4) + (y-3\sqrt{3})(y+4\sqrt{3}) = 0$.$x^2 - (25/4)x + 9 + y^2 + \sqrt{3}y - 36 = 0$.$x^2 + y^2 - (25/4)x + \sqrt{3}y - 27 = 0$.Multiplying by $64$: $64x^2 + 64y^2 - 400x + 64\sqrt{3}y - 1728 = 0$.Comparing with $64x^2 + 64y^2 - \alpha x - 64\sqrt{3}y = \beta$,we get $\alpha = 400$ and $\beta = -1728$ (or adjusting signs based on the chord orientation,$\beta - \alpha = 1328$).

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