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(B) The parabola is $y^2 = 2px$. Comparing this with $y^2 = 4ax$,we get $4a = 2p$,so $a = \frac{p}{2}$.The focus of the parabola is $S = (a, 0) = (\fr

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$(\frac{p}{2}, -p)$

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(B) The parabola is $y^2 = 2px$. Comparing this with $y^2 = 4ax$,we get $4a = 2p$,so $a = \frac{p}{2}$.The focus of the parabola is $S = (a, 0) = (\frac{p}{2}, 0)$.The directrix of the parabola is $x = -a = -\frac{p}{2}$.The circle is centered at $(\frac{p}{2}, 0)$ and touches the directrix $x = -\frac{p}{2}$.The radius $r$ of the circle is the distance from the focus to the directrix,which is $r = \frac{p}{2} - (-\frac{p}{2}) = p$.The equation of the circle is $(x - \frac{p}{2})^2 + y^2 = p^2$.Substitute $y^2 = 2px$ into the circle equation:$(x - \frac{p}{2})^2 + 2px = p^2$$x^2 - px + \frac{p^2}{4} + 2px = p^2$$x^2 + px - \frac{3p^2}{4} = 0$Using the quadratic formula $x = \frac{-p \pm \sqrt{p^2 - 4(1)(-\frac{3p^2}{4})}}{2} = \frac{-p \pm \sqrt{p^2 + 3p^2}}{2} = \frac{-p \pm 2p}{2}$.Since $x$ must be positive for the parabola $y^2 = 2px$,we take $x = \frac{p}{2}$.If $x = \frac{p}{2}$,then $y^2 = 2p(\frac{p}{2}) = p^2$,so $y = \pm p$.Thus,the points of intersection are $(\frac{p}{2}, p)$ and $(\frac{p}{2}, -p)$.

$A$ circle is drawn with its centre at the focus of the parabola $y^2 = 2px$ such that it touches the directrix of the parabola. Then a point of intersection of the circle and the parabola is

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