If Pleft(frac{1}{2}, 4right) and Q are the en | vedclass

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(C) The equation of the parabola is $y^2 = 32x$. Comparing this with $y^2 = 4ax$,we get $4a = 32$,so $a = 8$. The focus $S$ is $(8, 0)$.For a point $P

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$136$

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(C) The equation of the parabola is $y^2 = 32x$. Comparing this with $y^2 = 4ax$,we get $4a = 32$,so $a = 8$. The focus $S$ is $(8, 0)$.For a point $P(x_1, y_1)$ on the parabola,the coordinates of the other end $Q(x_2, y_2)$ of the focal chord are given by $x_2 = \frac{a^2}{x_1}$ and $y_2 = \frac{-4a^2}{y_1}$.Given $P = (\frac{1}{2}, 4)$,we have $x_2 = \frac{8^2}{1/2} = \frac{64}{1/2} = 128$ and $y_2 = \frac{-4(8^2)}{4} = -64$.Thus,$Q = (128, -64)$.The length of the focal chord $PQ$ is the distance $SP + SQ$. Alternatively,we calculate $SQ$ using the distance formula:$SQ = \sqrt{(128 - 8)^2 + (-64 - 0)^2} = \sqrt{120^2 + (-64)^2} = \sqrt{14400 + 4096} = \sqrt{18496} = 136$.

If $P\left(\frac{1}{2}, 4\right)$ and $Q$ are the ends of a focal chord of the parabola $y^2=32x$ and $S$ is the focus of the parabola,then $SQ=$

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