The normal drawn at a point (2, -4) on the pa | vedclass

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(D) Given the parabola $y^2 = 8x$,we have $4a = 8$,so $a = 2$.For a point $(at^2, 2at)$ on the parabola,we have $(2, -4) = (2t^2, 4t)$,which gives $t

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$30$

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(D) Given the parabola $y^2 = 8x$,we have $4a = 8$,so $a = 2$.For a point $(at^2, 2at)$ on the parabola,we have $(2, -4) = (2t^2, 4t)$,which gives $t = -1$.The normal at parameter $t$ meets the parabola again at parameter $t_2 = -t - \frac{2}{t}$.Substituting $t = -1$,we get $t_2 = -(-1) - \frac{2}{-1} = 1 + 2 = 3$.The coordinates of the point $(\alpha, \beta)$ are $(at_2^2, 2at_2) = (2(3)^2, 2(2)(3)) = (18, 12)$.Thus,$\alpha + \beta = 18 + 12 = 30$.

The normal drawn at a point $(2, -4)$ on the parabola $y^2 = 8x$ cuts the same parabola again at $(\alpha, \beta)$. Then $\alpha + \beta =$

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