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(C) The equation of the parabola is $y^2 = 8x$,where $4a = 8$,so $a = 2$.Let the chord be $y = mx + c$. The points of intersection of the line $y = mx

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(C) The equation of the parabola is $y^2 = 8x$,where $4a = 8$,so $a = 2$.Let the chord be $y = mx + c$. The points of intersection of the line $y = mx + c$ and the parabola $y^2 = 8x$ are given by $(mx + c)^2 = 8x$,which simplifies to $m^2x^2 + (2mc - 8)x + c^2 = 0$.Let the roots be $x_1$ and $x_2$. Then $x_1 + x_2 = -\frac{2mc - 8}{m^2}$ and $x_1x_2 = \frac{c^2}{m^2}$.The mid-point of the chord is $(h, k) = (\frac{x_1 + x_2}{2}, m(\frac{x_1 + x_2}{2}) + c)$.Since the circle has radius $4$ and touches the axis of the parabola $(y = 0)$,the $y$-coordinate of the center must be $4$ or $-4$. Thus,$k = 4$ (assuming positive).The length of the chord is $L = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2} = \sqrt{(x_1 - x_2)^2 + m^2(x_1 - x_2)^2} = |x_1 - x_2| \sqrt{1 + m^2}$.Since the circle has diameter $L$,the radius is $R = \frac{L}{2} = 4$,so $L = 8$.Using the property of the chord of a parabola,the slope $m = \frac{2}{t_1 + t_2}$ and the mid-point $y$-coordinate $k = 2(t_1 + t_2) = 4$,which gives $t_1 + t_2 = 2$.Thus,$m = \frac{2}{2} = 1$.

$A$ circle of radius $4$,drawn on a chord of the parabola $y^2 = 8x$ as diameter,touches the axis of the parabola. Then,the slope of the chord is

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