Q. 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 axis of the parabola is
JamiaJamia 2009
Solution:
In $ \Delta ABC, $ we have $ \tan \theta -\frac{y}{x} $ ...(i)
In $ \Delta BCD, $ we have $ \tan (90{}^\circ -\theta )=\frac{y}{CD} $ $ \Rightarrow $ $ CD=y\,\tan \theta =\frac{{{y}^{2}}}{x} $ [using Eq. (i)] $ \Rightarrow $ $ CD=\frac{4ax}{x}=4a $
