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設B點座標為(xb,xb^2+3),P為(x0,y0) 則 2x0=xb+6 2y0=xb^2+3 由得xb=2x0-6,代入得 2y0=39-24 x0+4x0^2 化簡得y0=2x0^2-12x0+39/2 所以P的軌跡方程為y=2x^2-12x+39/2
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