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y²;=4x的焦點F(1,0),準線x=-1設A(x1,y1),B(x2,y2)利用抛物線的定義則|AF|=x1+1,|BF|=x2+1∴|AB|=x1+x2+2直線為y=tanθ(x-1)代入抛物線方程則tan²;θ(x-1)²;=4x即tan²;θx²;-(2tan²;θ+4…
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