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(C) Let the coordinates of point $B$ be $(x, y)$.Since $B$ lies on the parabola $y^2 = 4ax$,we have $y^2 = 4ax$.The slope of $AB$ is $m_1 = \frac{y}{x

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$4a$

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(C) Let the coordinates of point $B$ be $(x, y)$.Since $B$ lies on the parabola $y^2 = 4ax$,we have $y^2 = 4ax$.The slope of $AB$ is $m_1 = \frac{y}{x}$.Since $BC \perp AB$,the slope of $BC$ is $m_2 = -\frac{x}{y}$.The equation of line $BC$ passing through $(x, y)$ is $(Y - y) = -\frac{x}{y}(X - x)$.To find the point $C$ on the axis of the parabola (where $Y = 0$),we set $Y = 0$:$-y = -\frac{x}{y}(X - x) \implies y^2 = x(X - x) \implies 4ax = x(X - x) \implies 4a = X - x$.Thus,the projection of $BC$ on the axis is $DC = X - x = 4a$.

$AB$ is a chord of the parabola $y^2 = 4ax$ with one endpoint $A$ at the vertex of the parabola. $BC$ is drawn perpendicular to $AB$ and meets the axis of the parabola at $C$. What is the projection of $BC$ on the axis of the parabola?

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