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(C) The equation of the parabola is $y^2 = x$. Comparing this with $y^2 = 4ax$,we get $4a = 1$,so $a = \frac{1}{4}$.Any normal to the parabola $y^2 =

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$C > \frac{1}{2}$

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(C) The equation of the parabola is $y^2 = x$. Comparing this with $y^2 = 4ax$,we get $4a = 1$,so $a = \frac{1}{4}$.Any normal to the parabola $y^2 = 4ax$ is given by $y = mx - 2am - am^3$.Substituting $a = \frac{1}{4}$,the equation becomes $y = mx - \frac{m}{2} - \frac{m^3}{4}$.If this normal passes through the point $(C, 0)$,then $0 = mC - \frac{m}{2} - \frac{m^3}{4}$.This simplifies to $m(C - \frac{1}{2} - \frac{m^2}{4}) = 0$.One root is $m = 0$,which corresponds to the normal along the $x$-axis.For three distinct normals to be drawn,the quadratic equation $\frac{m^2}{4} = C - \frac{1}{2}$ must have two distinct non-zero real roots.This requires $C - \frac{1}{2} > 0$,which implies $C > \frac{1}{2}$.

From a point $(C, 0)$,three normals are drawn to the parabola $y^2=x$. Then,

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