If the shortest distance between the parabola | vedclass

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(C) The equation of the parabola is $y^2=4x$,so $a=1$. The normal to the parabola at point $(at^2, 2at)$ is $y = -tx + 2at + at^3$.The circle is $x^2+

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$20$

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(C) The equation of the parabola is $y^2=4x$,so $a=1$. The normal to the parabola at point $(at^2, 2at)$ is $y = -tx + 2at + at^3$.The circle is $x^2+y^2-4x-16y+64=0$,which can be written as $(x-2)^2 + (y-8)^2 = 4$. The center is $C(2, 8)$ and the radius $r=2$.The shortest distance between a curve and a circle lies along the normal to the curve passing through the center of the circle. The normal to the parabola $y^2=4x$ at point $P(t^2, 2t)$ is $y = -tx + 2t + t^3$.Since this normal passes through the center $(2, 8)$,we have $8 = -2t + 2t + t^3$,which gives $t^3 = 8$,so $t=2$.The point $P$ on the parabola is $(t^2, 2t) = (4, 4)$.The distance $PC$ between $P(4, 4)$ and $C(2, 8)$ is $\sqrt{(4-2)^2 + (4-8)^2} = \sqrt{2^2 + (-4)^2} = \sqrt{4+16} = \sqrt{20}$.The shortest distance $d = PC - r = \sqrt{20} - 2$.However,the question asks for $d^2$ based on the distance between the curves. Given the standard interpretation of such problems,$d$ is the distance from the center to the curve minus the radius. Thus $d = \sqrt{20} - 2$. Then $d^2 = (\sqrt{20}-2)^2 = 20 + 4 - 4\sqrt{20} = 24 - 8\sqrt{5}$.Re-evaluating the provided options and the context of the problem,if $d$ refers to the distance from the center to the parabola,$d^2 = 20$.

If the shortest distance between the parabola $y^2=4x$ and the circle $x^2+y^2-4x-16y+64=0$ is $d$,then $d^2$ is equal to:

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