设双曲线方程为x^2/a^2-y^2/b^2=1.A(-a,0),B(a,0).设点P(m,n),则点Q(m,-n),向量PB=(a-m,-n), 向量AQ=(m+a,-n),因为向量PB×向量AQ=向量0,所以(a-m)( m+a)+n*n=0,即a^2-m^2+n^2=0.m^2= a^2+n^2.因为点P(m,n)在双曲线上,所以m^2/a^2-n^2/b^2=1.将m^2= a^2+n^2代入上式可得:(a^2+n^2)/a^2-n^2/b^2=1.1+ n^2/a^2-n^2/b^2=1.n^2/a^2-n^2/b^2=0,所以a^2=b^2,又因c^2=a^2+b^2,所以c^2=2a^2,c/a=√2,即...
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