AB is a chord of the parabola $y^{2} = 4 a x$ with vertex at A. BC is drawn perpendicular to AB meeting the axis at C. The projection of BC on the x-axis is-

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Solution

The correct option is C
4a

Let B be $a t^{2}, 2 a t$

Slope of $A B = \frac{2}{t} .$ Equation of BC is
$y - 2 a t = - \frac{t}{2} (x - a t^{2}) .$
This meets y = 0 at C. Then,
$C = (4 a + a t^{2}, 0) a n d D = (a t^{2}, 0)$
$∴ D C = 4 a + a t^{2} - a t^{2} = 4 a .$

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