The normal to the parabola {y^2 = 8x} at the | vedclass

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(D) The equation of the parabola is ${y^2 = 8x}$,which is of the form ${y^2 = 4ax}$,so ${a = 2}$.The point $(2, 4)$ corresponds to the parameter ${t_1

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$\{18, -12\}$

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(D) The equation of the parabola is ${y^2 = 8x}$,which is of the form ${y^2 = 4ax}$,so ${a = 2}$.The point $(2, 4)$ corresponds to the parameter ${t_1}$ where ${at_1^2 = 2}$ and ${2at_1 = 4}$.Substituting ${a = 2}$,we get ${2t_1^2 = 2 \implies t_1^2 = 1 \implies t_1 = 1}$ (since ${y > 0}$).If the normal at ${t_1}$ meets the parabola at ${t_2}$,then ${t_2 = -t_1 - \frac{2}{t_1}}$.Substituting ${t_1 = 1}$,we get ${t_2 = -1 - \frac{2}{1} = -3}$.The coordinates of the point are $(at_2^2, 2at_2) = (2(-3)^2, 2(2)(-3)) = (18, -12)$.

The normal to the parabola ${y^2 = 8x}$ at the point $(2, 4)$ meets the parabola again at the point

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