The point(s) on the parabola y^2 = 4x which a | vedclass

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(C) The equation of the circle is $x^2 + y^2 - 24y + 128 = 0$. Completing the square for $y$,we get $x^2 + (y - 12)^2 = 16$. The center of the circle

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$(4, 4)$

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(C) The equation of the circle is $x^2 + y^2 - 24y + 128 = 0$. Completing the square for $y$,we get $x^2 + (y - 12)^2 = 16$. The center of the circle is $C(0, 12)$ and its radius is $r = 4$.Let a point on the parabola $y^2 = 4x$ be $P(t^2, 2t)$.The distance between the center $C(0, 12)$ and point $P(t^2, 2t)$ is minimized when the normal to the parabola at $P$ passes through the center $C$.The slope of the tangent at $P(t^2, 2t)$ is $\frac{dy}{dx} = \frac{2}{y} = \frac{2}{2t} = \frac{1}{t}$.Thus,the slope of the normal at $P$ is $-t$.The slope of the line segment $CP$ is $\frac{2t - 12}{t^2 - 0} = \frac{2t - 12}{t^2}$.Equating the slopes: $-t = \frac{2t - 12}{t^2}$ $\Rightarrow -t^3 = 2t - 12$ $\Rightarrow t^3 + 2t - 12 = 0$.Testing integer roots,for $t = 2$: $2^3 + 2(2) - 12 = 8 + 4 - 12 = 0$. So,$t = 2$ is a solution.For $t = 2$,the point $P$ is $(t^2, 2t) = (2^2, 2(2)) = (4, 4)$.Thus,the point on the parabola closest to the circle is $(4, 4)$.

The point$(s)$ on the parabola $y^2 = 4x$ which are closest to the circle $x^2 + y^2 - 24y + 128 = 0$ is/are:

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