At what point on the parabola y^2 = 4x does t | vedclass

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(D) The equation of a normal to the parabola $y^2 = 4ax$ at the point $(am^2, -2am)$ is $y = mx - 2am - am^3$.For the parabola $y^2 = 4x$,we have $a =

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$(1, -2)$

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(D) The equation of a normal to the parabola $y^2 = 4ax$ at the point $(am^2, -2am)$ is $y = mx - 2am - am^3$.For the parabola $y^2 = 4x$,we have $a = 1$. Thus,the normal at $(m^2, -2m)$ is $y = mx - 2m - m^3$.The slope of this normal is $m$.If the normal makes equal angles with the coordinate axes,its inclination $	heta$ must be $45^{\circ}$ or $135^{\circ}$,so the slope $m = 	an(45^{\circ}) = 1$ or $m = 	an(135^{\circ}) = -1$.Case $1$: If $m = 1$,the point is $(1^2, -2(1)) = (1, -2)$.Case $2$: If $m = -1$,the point is $((-1)^2, -2(-1)) = (1, 2)$.Comparing with the given options,the correct point is $(1, -2)$.

At what point on the parabola $y^2 = 4x$ does the normal make equal angles with the coordinate axes?

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