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 ka. What's the value of 'k'?

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Solution

Here we are asked to find the projection of BC on the x-axis.

This is equal to the difference in the x coordinates of points B and C.

$L e t B \equiv (a t^{2}, 2 a t)$

Lets first get the equation of BC

Slope of $A B = \frac{2 a t - 0}{a t^{2} - 0} = \frac{2}{t}$

$∴$ Equation of BC is,

$y - 2 a t = (- \frac{t}{2}) (x - a t^{2})$

for getting point C. solve BC with x-axis

$2 a t = [\frac{t}{2}] . (x - a t^{2})$

$4 a = x - a t^{2}$

$x = 4 a + a t^{2}$

$C \equiv (4 a + a t^{2}, 0)$

$∴$ projection of BC on x-axis is $= 4 a + a t^{2} - a t^{2}$

$k a = 4 a$

$∴ k = 4$

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