Let the point P be the vertex of the parabola | vedclass

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(B) The equation of the parabola is $y = (x-3)^2 + 3$,so the vertex $P$ is $(3, 3)$.The equation of the circle is $(x-1)^2 + (y-2)^2 = 2$,with center

Vedclass · Accepted Answer

$20$

Vedclass · Answer

(B) The equation of the parabola is $y = (x-3)^2 + 3$,so the vertex $P$ is $(3, 3)$.The equation of the circle is $(x-1)^2 + (y-2)^2 = 2$,with center $O(1, 2)$ and radius $r = \sqrt{2}$.The distance $PO = \sqrt{(3-1)^2 + (3-2)^2} = \sqrt{4+1} = \sqrt{5}$.Let the line through $P$ make an angle $	heta$ with the line $PO$. Let $d$ be the perpendicular distance from $O$ to the line $RS$. Then $d = PO \sin 	heta = \sqrt{5} \sin 	heta$.The lengths $PR$ and $PS$ are the roots of the quadratic equation obtained by substituting the line equation into the circle equation. If $d_1, d_2$ are the distances $PR$ and $PS$,then $d_1 + d_2 = 2 \sqrt{r^2 - d^2} \cos \phi$ is not the standard approach. Instead,using the property of the chord,$PR + PS = 2 \sqrt{PO^2 - d^2} \cos \alpha$ where $\alpha$ is the angle between $PO$ and the line. More simply,$PR$ and $PS$ are $d \pm \sqrt{r^2 - d^2}$. Thus $PR+PS = 2 \sqrt{PO^2 - d^2} = 2 \sqrt{5 - d^2}$.To maximize $(PR+PS)^2 = 4(5 - d^2)$,we minimize $d^2$. The minimum value of $d$ is $0$ (when the line passes through the center $O$).Thus,the maximum value is $4(5 - 0) = 20$.

Let the point $P$ be the vertex of the parabola $y = x^2 - 6x + 12$. If a line passing through the point $P$ intersects the circle $x^2 + y^2 - 2x - 4y + 3 = 0$ at the points $R$ and $S$,then the maximum value of $(PR + PS)^2$ is:

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