If L is the normal drawn to the parabola y^2 | vedclass

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(D) The equation of the parabola is $y^2 = 8x$,so $4a = 8$,which gives $a = 2$.The point on the parabola at parameter $t$ is $(at^2, 2at) = (2t^2, 4t)

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$(3, \sqrt{2})$

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(D) The equation of the parabola is $y^2 = 8x$,so $4a = 8$,which gives $a = 2$.The point on the parabola at parameter $t$ is $(at^2, 2at) = (2t^2, 4t)$.For $t = \frac{1}{\sqrt{2}}$,the point is $(2(\frac{1}{2}), 4(\frac{1}{\sqrt{2}})) = (1, 2\sqrt{2})$.The equation of the normal to the parabola $y^2 = 4ax$ at parameter $t$ is $y + tx = 2at + at^3$.Substituting $a = 2$ and $t = \frac{1}{\sqrt{2}}$,the equation of the normal $L$ is $y + \frac{1}{\sqrt{2}}x = 2(2)(\frac{1}{\sqrt{2}}) + 2(\frac{1}{2\sqrt{2}}) = \frac{4}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \frac{5}{\sqrt{2}}$.Multiplying by $\sqrt{2}$,we get $x + \sqrt{2}y = 5$.The focus of the parabola is $(a, 0) = (2, 0)$.The foot of the perpendicular $(h, k)$ from $(x_1, y_1)$ to the line $Ax + By + C = 0$ is given by $\frac{h - x_1}{A} = \frac{k - y_1}{B} = -\frac{Ax_1 + By_1 + C}{A^2 + B^2}$.Here $A = 1, B = \sqrt{2}, C = -5$ and $(x_1, y_1) = (2, 0)$.$\frac{h - 2}{1} = \frac{k - 0}{\sqrt{2}} = -\frac{1(2) + \sqrt{2}(0) - 5}{1^2 + (\sqrt{2})^2} = -\frac{2 - 5}{1 + 2} = -\frac{-3}{3} = 1$.So,$h - 2 = 1 \implies h = 3$ and $\frac{k}{\sqrt{2}} = 1 \implies k = \sqrt{2}$.The foot of the perpendicular is $(3, \sqrt{2})$.

If $L$ is the normal drawn to the parabola $y^2 = 8x$ at the point $t = \frac{1}{\sqrt{2}}$,then the foot of the perpendicular drawn from the focus of the parabola on to the normal $L$ is

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