Question

已知抛物线y=x2+mx-3/4m2（m〉0）与x轴交于A，B两点

Answer 1

（1）y=x^2+mx-3/4m^2
=x^2+mx+1/4m^2-m^2
=(x+1/2m)^2-m^2
抛物线对称轴搏友为x=-1/2m
由m>0知-1/陵闭2m<0,即对称轴在y轴左侧
(2)y=x^2+mx-3/4m^2
=(x+3/2m)(x-1/2m)
由1/OB-1/OA=2/3>0可知，Xa<Xb
即A(-3/2m,0), B(1/2m,0)
带入1/OB-1/OA=2/3得 2/m+2/3m=2/3
解得m=4,则抛物线解析基汪槐式为y=x^2+4x-3
(3)由题意知C(0,-3/4m^2)
由△ABC是直角三角形可知（抱歉画不了图，自己画一下看看）△AOC∽△COB
从而列出比例式OA/OC=OC/OB
代入OA=3/2m,OB=1/2m,OC=3/4m^2,化简得
m^2=4/3,解得m=2√3/3(舍负)
则OA=√3,OB=√3/3,OC=1
∴S△ABC=1/2AB*OC=1/2*(4√3/3)*1=2√3/3

Answer 2

(2)
y=x^2+mx-3/4m^2
=(x+3/2m)(x-1/2m)
由1/OB-1/OA=2/3>0可知，Xa<Xb
即A(-3/2m,0), B(1/2m,0)
1/OB-1/OA=OA-OB/OB*OA=2/睁蠢3
由韦达定理得x1+x2=-m x1*x2=-3/4m^2
所以OB-OA= m OB*OA=3/4m^2
代入得m / 3/4m^2=2/3
m1=0 (舍去) m2=2
所悉埋陪以y=x^2+2x-3

(3)
令x=0则y=-3/4m^2
所以c(0,-3/4m^2)
co=3/4m^2
由射影定理得oc^2=OB*OA
(3/4m^2)^2=3/4m^2
m1=0(舍液源去) m2=2√3/3
OA=√3
OB=√3/3
OC=1
S△ABC=1/2AB*OC=(√3+√3/3)*1*1/2=2√3/3

Answer 3

（1）y=x^2+mx-3/4m^2
=x^2+mx+1/4m^2-m^2
=(x+1/2m)^2-m^2
抛物线对称轴为x=-1/2m
由m>0知-1/2m<0,即对称轴在y轴左侧
(2)y=x^2+mx-3/4m^2
=(x+3/2m)(x-1/2m)
由1/OB-1/OA=2/3>0可知，Xa<Xb
即A(-3/2m,0),B(1/2m,0)
带入1/OB-1/OA=2/3得2/m+2/3m=2/3
解得m=4,则抛物线基汪槐解析式为y=x^2+4x-3
(3)由题意知C(0,-3/4m^2)
由△ABC是直角三角形可知（抱歉画不了图，自己画一下看看）陵闭△AOC∽△COB
从而列出比例式OA/OC=OC/OB
代入OA=3/2m,OB=1/2m,OC=3/4m^2,化简得
m^2=4/3,解得m=2√3/搏友3(舍负)
则OA=√3,OB=√3/3,OC=1
∴S△ABC=1/2AB*OC=1/2*(4√3/3)*1=2√3/3