Suppose AB is a focal chord of the parabola y | vedclass

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(B) For the parabola $y^2 = 4ax$,we have $4a = 12$,so $a = 3$. The focus $S$ is $(3, 0)$.Let the focal chord $AB$ make an angle $	heta$ with the $x$-

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$108$

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(B) For the parabola $y^2 = 4ax$,we have $4a = 12$,so $a = 3$. The focus $S$ is $(3, 0)$.Let the focal chord $AB$ make an angle $	heta$ with the $x$-axis. The length of the focal chord is given by $l = 4a \operatorname{cosec}^2 	heta = 12 \operatorname{cosec}^2 	heta$.The distance $d$ of the chord from the origin $(0, 0)$ is the perpendicular distance from the origin to the line passing through $(3, 0)$ with slope $m = 	an 	heta$. The equation of the line is $y - 0 = 	an 	heta (x - 3)$,which is $x \sin 	heta - y \cos 	heta - 3 \sin 	heta = 0$.The distance $d = \frac{|0 \cdot \sin 	heta - 0 \cdot \cos 	heta - 3 \sin 	heta|}{\sqrt{\sin^2 	heta + \cos^2 	heta}} = |3 \sin 	heta| = 3 \sin 	heta$.Thus,$d^2 = 9 \sin^2 	heta$,which implies $\sin^2 	heta = \frac{d^2}{9}$.Substituting this into the expression for $l$: $l = 12 \cdot \frac{1}{\sin^2 	heta} = 12 \cdot \frac{9}{d^2} = \frac{108}{d^2}$.Therefore,$l \cdot d^2 = 108$.

Suppose $AB$ is a focal chord of the parabola $y^2=12x$ of length $l$ and slope $m < \sqrt{3}$. If the distance of the chord $AB$ from the origin is $d$,then $l \cdot d^2$ is equal to ....................

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