A beam of light travels along the line y = -4 | vedclass

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(B) The focus is $(2, 0)$ and the directrix is $x = -2$. The vertex is the midpoint of the focus and the intersection of the axis with the directrix,w

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$(2, 4)$

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(B) The focus is $(2, 0)$ and the directrix is $x = -2$. The vertex is the midpoint of the focus and the intersection of the axis with the directrix,which is $(0, 0)$. The distance $a$ from the vertex to the focus is $2$. The equation of the parabola is $y^2 = 4ax = 8x$.The incident beam travels along $y = -4$. Substituting $y = -4$ into $y^2 = 8x$,we get $(-4)^2 = 8x$,so $16 = 8x$,which gives $x = 2$. Thus,point $P$ is $(2, -4)$.According to the reflection property of a parabola,any light ray traveling parallel to the axis of the parabola will reflect off the parabola and pass through its focus. The axis of this parabola is the $x$-axis $(y = 0)$. The incident ray $y = -4$ is parallel to the $x$-axis.Therefore,the reflected ray passes through the focus $S(2, 0)$. The reflected ray is a vertical line passing through $P(2, -4)$ and $S(2, 0)$,which is the line $x = 2$.To find the point $Q$ where the reflected ray hits the parabola again,we substitute $x = 2$ into the parabola equation $y^2 = 8x$:$y^2 = 8(2) = 16$$y = \pm 4$.Since $P$ is $(2, -4)$,the other point $Q$ is $(2, 4)$.

$A$ beam of light travels along the line $y = -4$ from right to left and strikes a parabola at point $P$. If the focus and directrix of the parabola are $(2, 0)$ and $x = -2$ respectively,then find the coordinates of the point where the reflected beam contacts the parabola again.

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