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$\mathrm{AB}$ is a chord of a parabola $\mathrm{y}^2=4 \mathrm{ax},(\mathrm{a}>0)$ with vertex $\mathrm{A} . \mathrm{BC}$ is drawn perpendicular to $\mathrm{AB}$ meeting the axis at $\mathrm{C}$. The projection of $\mathrm{BC}$ on the axis of the parabola is
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The correct answer is:
4a unit
$$
\begin{aligned}
&\mathrm{B}\left(a \mathrm{at}^2, 2 a t\right) \therefore \mathrm{D}\left(\mathrm{at}^2, 0\right) \\
&\mathrm{m}_{\mathrm{AB}}=\frac{2 a \mathrm{at}^2}{\mathrm{at}^2}=\frac{2}{\mathrm{t}} \therefore \mathrm{m}_{\mathrm{BC}}=-\frac{\mathrm{t}}{2}
\end{aligned}
$$
$\therefore y-2 a t=-\frac{t}{2}\left(x-a t^2\right)$ [Equation of $B C$ ]
for $y=0,-2 a t=-\frac{t}{2}\left(x-a t^2\right)=-\frac{t}{2} x+\frac{a t^3}{2}$
$$
\begin{aligned}
&\Rightarrow \mathrm{tx}=4 a \mathrm{at}+\mathrm{at}^3 \\
&\Rightarrow \mathrm{x}=4 \mathrm{a}+\mathrm{at}^2 \\
&\therefore \mathrm{c}\left(4 \mathrm{a}+\mathrm{at}^2, 0\right)
\end{aligned}
$$
$\therefore$ Projection of $B C$ on the axis is $D C=A C-A D=4 a+a t^2-a t^2=4 a$
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