The number of points on the parabola y^2=x at | vedclass

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(C) The equation of the parabola is $y^2=x$,which is of the form $y^2=4ax$,so $4a=1$ or $a=\frac{1}{4}$.Any point on the parabola can be represented a

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$2$

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(C) The equation of the parabola is $y^2=x$,which is of the form $y^2=4ax$,so $4a=1$ or $a=\frac{1}{4}$.Any point on the parabola can be represented as $(at^2, 2at) = (\frac{t^2}{4}, \frac{t}{2})$.The slope of the tangent at this point is $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{1/2}{t/2} = \frac{1}{t}$.The slope of the normal at this point is $m_n = -\frac{1}{	ext{slope of tangent}} = -t$.According to the problem,the slope of the normal is equal to the $x$-coordinate of the point:$-t = \frac{t^2}{4}$.Rearranging gives $t^2 + 4t = 0$,which implies $t(t+4) = 0$.Thus,$t=0$ or $t=-4$.These two values of $t$ correspond to two distinct points on the parabola.Therefore,there are $2$ such points.

The number of points on the parabola $y^2=x$ at which the slope of the normal drawn at the point is equal to the $x$-coordinate of that point is

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