If the line 2x + 3y + n = 0 is a tangent to t | vedclass

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(C) The equation of a tangent to the parabola $y^2 = 4ax$ is $y = mx + \frac{a}{m}$.Given $y^2 = 8x$,we have $a = 2$,so the tangent is $y = mx + \frac

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$3x + y - 66 = 0$

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(C) The equation of a tangent to the parabola $y^2 = 4ax$ is $y = mx + \frac{a}{m}$.Given $y^2 = 8x$,we have $a = 2$,so the tangent is $y = mx + \frac{2}{m}$.Rewriting the given line $2x + 3y + n = 0$ as $y = -\frac{2}{3}x - \frac{n}{3}$.Comparing the slopes,$m = -\frac{2}{3}$.Equating the intercepts,$-\frac{n}{3} = \frac{2}{m} = \frac{2}{-2/3} = -3$,which gives $n = 9$.The point of contact is $(2n, 4\sqrt{n}) = (18, 12)$.For the parabola $y^2 = 8x$,the normal at point $(at^2, 2at)$ is $y = -tx + 2at + at^3$.Since $(at^2, 2at) = (18, 12)$ and $a = 2$,we have $2t^2 = 18$ $\Rightarrow t^2 = 9$ $\Rightarrow t = 3$ (taking positive root for the point).The equation of the normal is $y = -3x + 2(2)(3) + 2(3)^3 = -3x + 12 + 54$.Thus,$3x + y - 66 = 0$.

If the line $2x + 3y + n = 0$ is a tangent to the parabola $y^2 = 8x$,then the equation of the normal drawn at the point $(2n, 4\sqrt{n})$ to the parabola $y^2 = 8x$ is

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