Find the equation of the normal to the parabo | vedclass

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Question

(C) The equation of the normal to the parabola $y^2 = 4ax$ with slope $m$ is given by $y = mx - 2am - am^3$.Here,$a = 1$ and the normal passes through

Vedclass · Accepted Answer

$y = -x + 3$

Vedclass · Answer

(C) The equation of the normal to the parabola $y^2 = 4ax$ with slope $m$ is given by $y = mx - 2am - am^3$.Here,$a = 1$ and the normal passes through the point $(3, 0)$.Substituting these values into the equation: $0 = 3m - 2(1)m - (1)m^3$.This simplifies to $0 = m - m^3$,or $m(1 - m^2) = 0$.Thus,$m = 0$ or $m = \pm 1$.For $m = 0$,the equation is $y = 0$.For $m = 1$,the equation is $y = x - 2(1) - 1(1)^3 = x - 3$.For $m = -1$,the equation is $y = -x - 2(1)(-1) - 1(-1)^3 = -x + 2 + 1 = -x + 3$.Both $y = x - 3$ and $y = -x + 3$ are valid normals. Given the options,$y = -x + 3$ is provided as option $C$.

Find the equation of the normal to the parabola $y^2 = 4x$ passing through the point $(3, 0)$.

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