Three normals to the parabola y^2 = x are dra | vedclass

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(C) The equation of the normal to the parabola $y^2 = 4ax$ in slope form is $y = mx - 2am - am^3$.For the given parabola $y^2 = x$,we have $4a = 1$,wh

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

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(C) The equation of the normal to the parabola $y^2 = 4ax$ in slope form is $y = mx - 2am - am^3$.For the given parabola $y^2 = x$,we have $4a = 1$,which implies $a = \frac{1}{4}$.Substituting $a$ into the normal equation,we get $y = mx - \frac{1}{2}m - \frac{1}{4}m^3$.Since the normal passes through the point $(C, 0)$,we substitute $x = C$ and $y = 0$:$0 = mC - \frac{1}{2}m - \frac{1}{4}m^3$.This simplifies to $m(C - \frac{1}{2} - \frac{1}{4}m^2) = 0$.One solution is $m = 0$. For the other two normals to be real and distinct,the quadratic equation $C - \frac{1}{2} - \frac{1}{4}m^2 = 0$ must have two distinct real roots for $m^2$.This requires $C - \frac{1}{2} > 0$,which implies $C > \frac{1}{2}$.

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

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