已知抛物线y^2=4x的焦点为f,直线l交抛物线于a,b两点,若点a,b的横坐标之和为8,试证明:线段ab的垂直平

分线过定点。
2025-03-16 17:16:00
推荐回答(3个)
回答1:

设直线l的方程为:y=kx+b(k≠0),代入到y^2=4x,
得:(kx+b)^2=4x,即k^2x^2+(2kb-4)x+b^2=0.
设A(x1,y1),B(x2,y2),AB中点为C(x0,y0),所以有:
x1+x2=-(2kb-4)/k^2=8,整理得b=2/k-4k.
又y1+y2=k(x1+x2)+2b=8k+2b=4/k.所以有:
x0=4,y0=2/k.即C(4,2/k).所以线段AB的垂直平分线方程:
y-2/k=(-1/k)(x-4),即y=-1/k*x+6/k=-1/k(x-6).
所以线段AB的垂直平分线经过定点(6,0).

回答2:

设A(x1,y1),B(x2,y2),AB中点为C(x0,y0),
则 x0=(x1+x2)/2=4,y0=(y1+y2)/2。
则 y1^2=4x1,y2^2=4x2,
两式相减得 (y1+y2)(y2-y1)=4(x2-x1),
因此,直线AB的斜率 k=(y2-y1)/(x2-x1)=4/(y1+y2)=2/y0,
所以,直线AB的垂直平分线的斜率k1=-1/k=-y0/2。
因此,AB的垂直平分线的方程为 y=-y0/2*(x-4)+y0,
显然,它恒过定点(6,0)。

回答3:

抛物线y^2=4x的焦点为f(1,0)
设直线l为x=ky+b,代入抛物线,得
y^2=4x=4ky+4b,即y^2-4ky-4b=0
∴y1+y2=4k,已知x1+x2=8
直线l斜率为1/k,则ab的垂直平分线斜率为-k,
ab中点坐标为((x1+x2)/2,(y1+y2)/2)=(4,2k)
∴中垂线方程为y-2k=-k*(x-4),即y=-kx+6k=-k(x-6)
∴中垂线过定点(6,0)

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