求值
过y^2=x/a(a>0)焦点F任作一直线交抛物线于P、Q,若线段PF、FQ长度分别为p、q,试求1/p+1/q的值.
解: 将抛物线化为极坐标方程: L=(1/2a)/(1-cost) 设P(L1,t)、Q(L2,t+兀),代入得到 L1=(1/2a)/(1-cost), L2=(1/2a)/(1-cos(t+兀))=(1/2a)/(1+cost) 故1/p+1/q =1/L1+1/L2 =2a(1-cost)+2a(1+cost) =4a. 即所求值为4a.
答:解:抛物线y=(nn+n)xx-(2n+1)x+1与x轴交于An,Bn两点 ==>(n^2+n)x^2-(2n+1)x+1=0 == >(nx-1)[(n+1)...详情>>
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