Let P be the point on the parabola y^2=4x whi | vedclass

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(B) The center of the circle $x^2+y^2-4x-16y+64=0$ is $S(2, 8)$ and its radius $r$ is $\sqrt{2^2+8^2-64} = 2$.Let $P = (t^2, 2t)$ be a point on the pa

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$A, C, D$

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(B) The center of the circle $x^2+y^2-4x-16y+64=0$ is $S(2, 8)$ and its radius $r$ is $\sqrt{2^2+8^2-64} = 2$.Let $P = (t^2, 2t)$ be a point on the parabola $y^2=4x$.For $P$ to be the closest point to $S$,the line $SP$ must be the normal to the parabola at $P$.The slope of the tangent at $P$ is $\frac{1}{t}$,so the slope of the normal is $-t$.The slope of $SP$ is $\frac{2t-8}{t^2-2}$.Setting $\frac{2t-8}{t^2-2} = -t$,we get $2t-8 = -t^3+2t$,which implies $t^3=8$,so $t=2$.Thus,$P = (4, 4)$.$(A)$ $SP = \sqrt{(4-2)^2+(4-8)^2} = \sqrt{2^2+(-4)^2} = \sqrt{20} = 2\sqrt{5}$. This is correct.$(C)$ The slope of the normal at $P(4, 4)$ is $-t = -2$. The equation of the normal is $y-4 = -2(x-4)$,or $y = -2x+12$. The $x$-intercept is found by setting $y=0$,giving $2x=12$,so $x=6$. This is correct.$(D)$ Since $SP$ is the normal to the parabola at $P$,and $S$ is the center of the circle,the line $SP$ passes through the center of the circle. Thus,$SP$ is a normal to the circle at $Q$. The slope of $SP$ is $-2$. The tangent at $Q$ is perpendicular to the normal $SQ$,so its slope is $-\frac{1}{-2} = \frac{1}{2}$. This is correct.$(B)$ $SQ$ is the radius of the circle,so $SQ=2$. $SP=2\sqrt{5}$. $QP = SP - SQ = 2\sqrt{5}-2$. Thus $SQ:QP = 2 : (2\sqrt{5}-2) = 1 : (\sqrt{5}-1) = (\sqrt{5}+1) : 4$. This is incorrect.Therefore,the correct options are $A, C, D$.

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