The minimum distance between the parabola y^2 | vedclass

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(C) The parabola is $y^2 = 8x$,so $4a = 8$,which gives $a = 2$. The general point on the parabola is $P(2t^2, 4t)$.The slope of the tangent at $P$ is

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$4\sqrt{2}$

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(C) The parabola is $y^2 = 8x$,so $4a = 8$,which gives $a = 2$. The general point on the parabola is $P(2t^2, 4t)$.The slope of the tangent at $P$ is given by $\frac{dy}{dx} = \frac{4}{2t} = \frac{2}{t}$.For the minimum distance,the tangent at $P$ must be parallel to the line $x + y + 4 = 0$,which has a slope of $-1$.Setting $\frac{2}{t} = -1$,we get $t = -2$.The point $P$ is $(2(-2)^2, 4(-2)) = (8, -8)$.The distance from point $P(8, -8)$ to the line $x + y + 4 = 0$ is $d = \frac{|8 - 8 + 4|}{\sqrt{1^2 + 1^2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2}$.The minimum distance between the parabola and its image is twice the distance from the point $P$ to the line of reflection,which is $2 	imes (2\sqrt{2}) = 4\sqrt{2}$.

The minimum distance between the parabola $y^2 = 8x$ and its image with respect to the line $x + y + 4 = 0$ is:

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