
2个回答
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设交点A(x1,y1),B(x2,y2)
联立方程得y^2-px+p=0
yi+y2=p,y1*y2=p
(y1-y2)^2=(y1+y2)^2-4y1*y2=p^2-p
(x1-x2)^2=(y1-y2)^2/4
|AB|^2=(x1-x2)^2+(y1-y2)^2=5/4*(p^2-4p)=15
∴p=6或-2
y^2=12x,或y^2=-4x
联立方程得y^2-px+p=0
yi+y2=p,y1*y2=p
(y1-y2)^2=(y1+y2)^2-4y1*y2=p^2-p
(x1-x2)^2=(y1-y2)^2/4
|AB|^2=(x1-x2)^2+(y1-y2)^2=5/4*(p^2-4p)=15
∴p=6或-2
y^2=12x,或y^2=-4x
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