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

- a
- 2a
- 4a
- 8a

## The correct answer is: 4a

### Let B be (at^{2}, 2at)

Slope of AB =.

Equation of BC is y – 2at = – (x – at^{2}).

This meets y = 0 at C whose x-coordinate = 4a + at^{2}

D = (at^{2}, 0)

DC = 4a + at^{2} – at^{2} = 4a.

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