Let LL^{prime} be the latus rectum and PQ be | vedclass

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(D) The given parabola is $y^2=16x$,which is of the form $y^2=4ax$,so $a=4$.The focus $S$ of the parabola is $(a, 0) = (4, 0)$.The coordinates of the

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$12\sqrt{5}$

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(D) The given parabola is $y^2=16x$,which is of the form $y^2=4ax$,so $a=4$.The focus $S$ of the parabola is $(a, 0) = (4, 0)$.The coordinates of the endpoints of the latus rectum $LL^{\prime}$ are $(a, 2a)$ and $(a, -2a)$,so $L=(4, 8)$ and $L^{\prime}=(4, -8)$.Since $PQ$ is a focal chord passing through $P(1, 4)$ and $S(4, 0)$,the slope of $PQ$ is $m = \frac{0-4}{4-1} = -\frac{4}{3}$.The equation of the line $PQ$ is $y - 0 = -\frac{4}{3}(x - 4)$,which simplifies to $4x + 3y - 16 = 0$.To find the coordinates of $Q$,we substitute $x = \frac{y^2}{16}$ into the line equation: $4(\frac{y^2}{16}) + 3y - 16 = 0$ $\Rightarrow \frac{y^2}{4} + 3y - 16 = 0$ $\Rightarrow y^2 + 12y - 64 = 0$.Factoring gives $(y+16)(y-4) = 0$,so $y=4$ (point $P$) or $y=-16$ (point $Q$).For $y=-16$,$x = \frac{(-16)^2}{16} = 16$,so $Q=(16, -16)$.Now,the distance $LQ = \sqrt{(16-4)^2 + (-16-8)^2} = \sqrt{12^2 + (-24)^2} = \sqrt{144 + 576} = \sqrt{720} = 12\sqrt{5}$.

Let $LL^{\prime}$ be the latus rectum and $PQ$ be the focal chord of the parabola $y^2=16x$. If $P=(1,4)$ and $P, L$ lie in the same quadrant,then $LQ=$

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