Find the point of intersection of the parabol | vedclass

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(A) The parabola is given by $y^2 = 2px$. The focus of this parabola is $S = (p/2, 0)$.The directrix of the parabola is $x = -p/2$,or $x + p/2 = 0$.Th

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$(p/2, p)$

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(A) The parabola is given by $y^2 = 2px$. The focus of this parabola is $S = (p/2, 0)$.The directrix of the parabola is $x = -p/2$,or $x + p/2 = 0$.The circle has its center at $(p/2, 0)$ and touches the directrix. The radius $r$ of the circle is the distance from the focus $(p/2, 0)$ to the directrix $x = -p/2$,which is $r = |p/2 - (-p/2)| = |p| = p$.The equation of the circle is $(x - p/2)^2 + y^2 = p^2$.Substitute $y^2 = 2px$ into the circle equation:$(x - p/2)^2 + 2px = p^2$$x^2 - px + p^2/4 + 2px = p^2$$x^2 + px - 3p^2/4 = 0$Solving for $x$ using the quadratic formula:$x = \frac{-p \pm \sqrt{p^2 - 4(1)(-3p^2/4)}}{2} = \frac{-p \pm \sqrt{p^2 + 3p^2}}{2} = \frac{-p \pm 2p}{2}$So,$x = p/2$ or $x = -3p/2$.If $x = p/2$,$y^2 = 2p(p/2) = p^2$,so $y = \pm p$. The points are $(p/2, p)$ and $(p/2, -p)$.If $x = -3p/2$,$y^2 = 2p(-3p/2) = -3p^2$,which has no real solution for $y$.Thus,the points of intersection are $(p/2, p)$ and $(p/2, -p)$.

Find the point of intersection of the parabola $y^2 = 2px$ and a circle whose center is at the focus of the parabola and which touches the directrix of the parabola.

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