If the normal drawn at P(8, 16) to the parabo | vedclass

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Question

(A) The equation of the parabola is $y^2 = 32x$,so $4a = 32$,which implies $a = 8$. Point $P$ is $(8, 16)$,which corresponds to $(at^2, 2at) = (8t^2,

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$x + 3y + 72 = 0$

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(A) The equation of the parabola is $y^2 = 32x$,so $4a = 32$,which implies $a = 8$. Point $P$ is $(8, 16)$,which corresponds to $(at^2, 2at) = (8t^2, 16t)$. Thus,$t_1 = 2$. The normal at $t_1$ meets the parabola at $t_2 = -t_1 - \frac{2}{t_1} = -2 - \frac{2}{2} = -3$. The coordinates of $Q$ are $(at_2^2, 2at_2) = (8(-3)^2, 16(-3)) = (72, -48)$. The equation of the tangent at $Q(x_1, y_1)$ to $y^2 = 4ax$ is $yy_1 = 2a(x + x_1)$. Substituting $y_1 = -48$,$a = 8$,and $x_1 = 72$: $y(-48) = 2(8)(x + 72)$ $-48y = 16(x + 72)$ $-3y = x + 72$ $x + 3y + 72 = 0$.

If the normal drawn at $P(8, 16)$ to the parabola $y^2 = 32x$ meets the parabola again at $Q$,then the equation of the tangent drawn at $Q$ to the parabola is

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