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(C) The equation of the given parabola is $y^2 = 2px$. The focus is $F\left(\frac{p}{2}, 0\right)$ and the directrix is $x = -\frac{p}{2}$.Since the c

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

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(C) The equation of the given parabola is $y^2 = 2px$. The focus is $F\left(\frac{p}{2}, 0\right)$ and the directrix is $x = -\frac{p}{2}$.Since the circle is centered at the focus and touches the directrix,its radius $r$ 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$$(x + \frac{3p}{2})(x - \frac{p}{2}) = 0$.Since $x = -\frac{3p}{2}$ is outside the parabola's domain for $y^2 = 2px$ (assuming $p > 0$),we take $x = \frac{p}{2}$.Substituting $x = \frac{p}{2}$ into $y^2 = 2px$,we get $y^2 = 2p(\frac{p}{2}) = p^2$,so $y = \pm p$.The points of intersection are $\left(\frac{p}{2}, p\right)$ and $\left(\frac{p}{2}, -p\right)$.

If a circle with its centre at the focus of the parabola $y^2 = 2px$ is 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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