直线为：x+y=2；原点到直线的距离=根号2设抛物线：x^2=ay。|pq|=根号（2a^2+8a)s=3所以 a=（根号13)-2

解:设抛物线方程为y=ax^2 (a>0)经过点A(2,0)和(0,2)的直线方程:y=-x+2设P(x1,y1),Q(x2,y2),x1<0,x2>0三角形OPQ的面积=(y1+y2)(x2-x1)/2-y1(-x1)/2-y2*x2/2=3即(y1+y2)(x2-x1)+y1x1-y2*x2=6y=-x+2y=ax^2 即ax^2+x-2=0x1+x2=-1/a,x1*x2=-2/ay1+y2=-(x1+x2)+4=1/a+4x2-x1=√[(x1+x2)^2-4x1*x2]=√(1+8a)/ax1y1-x2y2=x1(-x1+2)-x2(-x2+2)=(x2-x1)(x2+x1-2)=[√(1+8a)/a](-1/a-2)所以(1/a+4)(√(1+8a)/a)+[√(1+8a)/a](-1/a-2)=6整理得:9a^2-8a-1=0解得a=1或a=-1/9(舍去)所以抛物线的表达式为y=x^2