直线y=ax+1与椭圆3x^2+y^2=2相交与P、Q两点。当a为何值时以PQ为直径的圆过坐标原点！把y=ax+1代入椭圆3x^2+y^2=2得:(3+a^2)

把y=ax+1代入椭圆3x^2+y^2=2得:
(3+a^2)x^2+2ax-1=0
△=4a^2+4(3+a^2)>0;
设P(x1,y1),Q(x2,y2)
则有x1+x2=-2a/(a^2+3),x1*x2=-1/(a^2+3),
所以y1y2=(ax1+1)(ax2+1)=a^2x1x2+a(x1+x2)+1
又0P⊥OQ,所以向量OP*向量OQ=x1x2+y1y2=0,
即:(a^2+1)x1x2+a(x1+x2)+1=0
所以:-(a^2+1)/(a^2+3)-2a^2/(a^2+3)+1=0,解得a=±1.

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