Question

过抛物线y=ax^2的焦点F，用一直线交抛物线于P,Q两点，若线段PF与PQ的长分别为p,q，则1/p+1/q等于

Answer 1

焦点F(0,1/4a),准线y=-1/4a,设P(x1,y1),Q(x2,y2),直线PQ:x=ky+k/4a由抛物线第二定义, p=PF=y1+1/4a, q=PF2=y2+1/4a联立y=ax^2,x=ky+k/4a, 得16a^2k^2y^2+(8ak^2-16a)y+k^2=0∴大尺y1+y2=(16a-8ak^2)/16a^2k^2=(2-k^2)/2ak^2, y1y2=k^2/16a^2k^2=1/16a^21/p+1/q=1/(y1+1/4a)+1/(y2+1/4a)=[(y1+y2)+1/2a]/滚核高[y1y2+(y1+y2)/4a+1/16a^2] =[(2-k^2)/2ak^2+1/2a]/[1/氏含16a^2+(2-k^2)/2ak^2/4a+1/16a^2](同乘8a^2k^2) =[4a(2-k^2)+4ak^2]/[k^2+2-k^2]=8a/2=4a