对于抛物线y^2=4x上任意一点Q,点P(a,0)满足绝对值PQ大于等于绝对值a,则a的取值范围

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点Q在抛物线y^2=4x上,设其坐标为:Q(y^2/4,y)|PQ| = |a| === |PQ|^2 = a^2 == (y^2/4 - a)^2 + (0 - y)^2 = a^2== (y^2)*(y^2 -8ay +16) = 0== y^2 -8ay +16 = 0若y取任意值,上式都成立,则:判别式 -1 < a < 1