作业帮设坐标原点为O,曲线y^2=2x与过点(1/2,0)的直线交于A、B两点,则向量OA×向量OB的值为
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设坐标原点为O,曲线y^2=2x与过点(1/2,0)的直线交于A、B两点,则向量OA×向量OB的值为
设坐标原点为O,曲线y^2=2x与过点(1/2,0)的直线交于A、B两点,则向量OA×向量OB的值为
设坐标原点为O,曲线y^2=2x与过点(1/2,0)的直线交于A、B两点,则向量OA×向量OB的值为
因为直线过点（0.5,0）,所以设直线为y=k(x-0.5)
设A（X1,Y1）,B(X2,Y2),那么：向量OA*向量OB=X1X2+Y1Y2
把直线的解析式代入曲线方程中,先消去y,得到：[k(x-0.5)]^2=2x
k^2x^2-(k^2+2)x+k^2/4=0
由伟达定理得到：X1X2=(k^2/4)/k^2=1/4
由直线解析式得到：x=y/k+0.5,再消去x得到：y^2=2y/k+1
即是：y^2-(2/k)y-1=0
由伟达定理得到：Y1Y2=-1
所以,向量OA*向量OB=-1+1/4=-3/4
回答完毕,
let the straight line L: through (1/2, 0) be
y /(x-1/2) = m
y = (2mx -m)/2
or x = (2y+m)/2m
L cuts y^2 = 2x at A,B
let A = ( x1, y1), B= ( x2, y2)
[(2mx-m)/2]^2 = 2x
m^2(2...
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let the straight line L: through (1/2, 0) be
y /(x-1/2) = m
y = (2mx -m)/2
or x = (2y+m)/2m
L cuts y^2 = 2x at A,B
let A = ( x1, y1), B= ( x2, y2)
[(2mx-m)/2]^2 = 2x
m^2(2x-1)^2 = 8x
4m^2x^2 - 4(m^2+2)x +m^2 =0
=> product of roots = x1x2 = m^2/(4m^2) =1/4
Also,
y^2 = 2x
y^2 = 2((2y+m)/2m)
my^2 = 2y+m
my^2-2y-m =0
=> y1y2 = -1
OA.OB
=(x1,y1).(x2,y2)
=x1x2+y1y2
= 1/4 -1
= -3/4
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