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令B(a,b),則有a^2-b^2+a+2b+3=0 AB的中點(a/2,(b-1)/2) x=a/2---> a=2x y=(b-1)/2---> b=2y+1 將a,b代入上式得:2x^2-(4y^2+4y+1)+2x+4y+2+3=0 化簡即得軌跡方程:x^2-2y^2+x+2=0
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