数学。!过抛物线y平方=x上一点A(4,2)作倾斜角互补的两条直线AB,AC交抛物线于B,C两点,求证:直线BC的斜率是定值。!
AB: y-2=k(x-4)
AC: y-2=-k(x-4)
联立
y²=x
y-2=k(x-4)
===>ky²-y-4k+2=0
根据韦达==> yB+2 =1/k ==>yB =1/k -2
带入y²=x ==>xB =(1-2k)²/k²
用-k代k
==>yC =-1/k -2
xC=(1+2k)²/k²
===>kBC =(yC-yB)/(xC-xB)
= (-2/k)/(8/k)
=-1/4是定值
。 全部
AB: y-2=k(x-4)
AC: y-2=-k(x-4)
联立
y²=x
y-2=k(x-4)
===>ky²-y-4k+2=0
根据韦达==> yB+2 =1/k ==>yB =1/k -2
带入y²=x ==>xB =(1-2k)²/k²
用-k代k
==>yC =-1/k -2
xC=(1+2k)²/k²
===>kBC =(yC-yB)/(xC-xB)
= (-2/k)/(8/k)
=-1/4是定值
。
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