The point on the parabola y^2 = 8x at which t | vedclass

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(B) The equation of the parabola is $y^2 = 8x$. Comparing with $y^2 = 4ax$,we get $a = 2$.The slope of the given line $x - 2y + 5 = 0$ is $m = 1/2$.Si

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$(1/2, -2)$

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(B) The equation of the parabola is $y^2 = 8x$. Comparing with $y^2 = 4ax$,we get $a = 2$.The slope of the given line $x - 2y + 5 = 0$ is $m = 1/2$.Since the normal is parallel to this line,the slope of the normal is $m_n = 1/2$.The slope of the normal to the parabola $y^2 = 4ax$ at point $(at^2, 2at)$ is $-t$.Equating the slopes: $-t = 1/2$,so $t = -1/2$.The coordinates of the point are $(at^2, 2at) = (2(-1/2)^2, 2(2)(-1/2)) = (2(1/4), -2) = (1/2, -2)$.Thus,the required point is $(1/2, -2)$.

The point on the parabola $y^2 = 8x$ at which the normal is parallel to the line $x - 2y + 5 = 0$ is

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