题目

抛物线y^2=6x上有两动点A（x1,y1）,B(x2,y2）,x1≠x2且x1+x2=4,AB的中垂线交x轴于C
求△ABC面积最大值

提问时间：2021-02-24

答案

x1+x2=4,(x1+x2)/2=4/2=2
AB中点P(2,a)
设直线AB:y-a=k(x-2)
x=(y+2k-a)/k
y^2=6x=6(y+2k-a)/k
ky^2-6y+6a-12k=0
y1+y2=6/k
a=(y1+y2)/2=3/k
y1*y2=(6a-12k)/k=(18-12k^2)/k^2
(y1-y2)^2=(y1+y2)^2-4y1*y2=(48k^2-36)/k^2
|y1-y2|=2√[(12k^2-9)/k^2]
直线CP:y-3/k=-(x-2)/k
y=0,xC=5
C(5,0)
设AB与X轴的交点D,则
AB:y-a=k(x-2)
y=0,xD=(2k^2-3)/k^2
|CD|=3(1+k^2)/k^2
△ABC面积=(1/2)*|CD|*|y1-y2|=s
s=(1/2)*[3(1+k^2)/k^2]*2√[(12k^2-9)/k^2]
=3(1/k^2+1)/k^2]*√(12-9/k^2)
高中知识不能解.
k无穷大时-9/k^2=0,1/k^2=0
s=3√12=6√3