过定点A(m,n)(n不等于0)的直线l交抛物线y=ax^2(a>0）与两点P、Q,若角POQ=90度，则l与y轴的交点是？
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设过定点A(m,n)(n不等于0)的直线l为：y-n = k(x-m)将y代入抛物线y=ax^2方程，得：ax^2 -kx+(km-n)=0直线l交抛物线于P(x1,y1),Q(x2,y2), l与y轴的交点R(0,n-km)== x1+x2 = k/a, x1*x2 = (km-n)/a角POQ=90度 == (y1/x1)*(y2/x2) = -10 = y1y2 +x1x2 = [k(x1-m)+n]*[k(x2-m)+n] + x1x2= k^2*x1x2+k(n-km)(x1+x2)+(n-km)^2 + x1x2== (1+k^2)*(km-n)/a +k(n-km)(k/a) +(n-km)^2 = 0== (n-km)(n-km-1) = 0n不等于0, == n-km-1 = 0n-km = 1==l与y轴的交点R(0,n-km) = R(0,1)。