已知直线y=ax+1与双曲线3x^2-y^2=1相交于点A,B两点,O为坐标原点,如果OA与OB垂直,求a的值假设A(X1,Y1),B(X2,Y2)X1X2+Y1Y2=0y=ax+1带入3x^2-y^2=1(3-a^2)x^2-2ax-2=0x1x2=-2/(3-a^2)我做到这里了(3-a^2)y^2-6ay+2a^2=0 y1y2=2a^2/(3-a)-2/(3-a^2)+2a^2/(3-a^2)=0a=±1我不明白这步(3-a^2)y^2-6ay+2a^2=0 是怎么得来的

已知直线y=ax+1与双曲线3x^2-y^2=1相交于点A,B两点,O为坐标原点,如果OA与OB垂直,求a的值
假设A(X1,Y1),B(X2,Y2)
X1X2+Y1Y2=0
y=ax+1带入3x^2-y^2=1
(3-a^2)x^2-2ax-2=0
x1x2=-2/(3-a^2)
我做到这里了
(3-a^2)y^2-6ay+2a^2=0
y1y2=2a^2/(3-a)
-2/(3-a^2)+2a^2/(3-a^2)=0
a=±1
我不明白这步(3-a^2)y^2-6ay+2a^2=0 是怎么得来的