Let P(alpha, beta) be a point on the parabola | vedclass

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(A) The equation of the circle is $x^2 - 4x + y^2 - 20y + 103 = 0$.Completing the square,we get $(x-2)^2 - 4 + (y-10)^2 - 100 + 103 = 0$,which simplif

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$20$

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(A) The equation of the circle is $x^2 - 4x + y^2 - 20y + 103 = 0$.Completing the square,we get $(x-2)^2 - 4 + (y-10)^2 - 100 + 103 = 0$,which simplifies to $(x-2)^2 + (y-10)^2 = 1$.The center of the circle is $C(2, 10)$ and its radius is $r = 1$.Let the point $P$ on the parabola $y^2 = 4x$ be $(t^2, 2t)$.The distance between $P$ and the center $C$ is minimized when the normal to the parabola at $P$ passes through the center $C(2, 10)$.The slope of the tangent at $P(t^2, 2t)$ is $\frac{dy}{dx} = \frac{2}{2t} = \frac{1}{t}$.The slope of the normal at $P$ is $-t$.The slope of the line segment $CP$ is $\frac{2t - 10}{t^2 - 2}$.Equating the slopes: $-t = \frac{2t - 10}{t^2 - 2}$.$-t^3 + 2t = 2t - 10$,which implies $t^3 = 10$,so $t = (10)^{1/3}$.Thus,$\alpha = t^2 = (10)^{2/3}$ and $\beta = 2t = 2(10)^{1/3}$.Therefore,$\alpha \beta = (10)^{2/3} 	imes 2(10)^{1/3} = 2(10)^{2/3 + 1/3} = 2(10)^1 = 20$.

Let $P(\alpha, \beta)$ be a point on the parabola $y^2 = 4x$ which is at minimum distance from the circle $x^2 + y^2 - 4x - 20y + 103 = 0$. Then $\alpha \beta$ is

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