AB is a chord of a parabola y^2 = 4ax, (a 0 | vedclass

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(D) Let the coordinates of point $B$ be $(at^2, 2at)$. The vertex $A$ is at $(0, 0)$.The slope of $AB$ is $m_{AB} = \frac{2at - 0}{at^2 - 0} = \frac{2

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$4a$ unit

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(D) Let the coordinates of point $B$ be $(at^2, 2at)$. The vertex $A$ is at $(0, 0)$.The slope of $AB$ is $m_{AB} = \frac{2at - 0}{at^2 - 0} = \frac{2}{t}$.Since $BC \perp AB$, the slope of $BC$ is $m_{BC} = -\frac{1}{m_{AB}} = -\frac{t}{2}$.The equation of line $BC$ passing through $B(at^2, 2at)$ with slope $-\frac{t}{2}$ is:$y - 2at = -\frac{t}{2}(x - at^2)$To find the point $C$ on the axis of the parabola $(y = 0)$:$0 - 2at = -\frac{t}{2}(x - at^2)$$4at = t(x - at^2)$$4a = x - at^2$$x = 4a + at^2$So, the coordinates of $C$ are $(4a + at^2, 0)$.The projection of $BC$ on the axis is the distance $DC$, where $D$ is the projection of $B$ on the axis, i.e.,$D(at^2, 0)$.$DC = |x_C - x_D| = |(4a + at^2) - at^2| = 4a$ units.

$AB$ is a chord of a parabola $y^2 = 4ax, (a > 0)$ with vertex $A$. $BC$ is drawn perpendicular to $AB$ meeting the axis at $C$. The projection of $BC$ on the axis of the parabola is

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