已知抛物线Y^2=2pX(p>0),焦点为F,O为顶点,Q为准线上一点,tan∠OQF=1/3求Q点坐标

P>0
F(0.5p,0)
准线x=-0.5p与X轴交于点A(-0.5p,0),设Q(-0.5p,y),则
AF=p,AO=0.5p
∠OQF=∠AOQ-∠AFQ
tan∠AOQ=|y/(0.5p)|=|2y/p|,tan∠AFQ=|y/p|
tan∠OQF=tan(∠AOQ-∠AFQ)=(|2y/p|-|y/p|)/(1+|2y/p|*|y/p|)=1/3
p|y|/(p^2+2y^2)=1/3
y=0.5p,p,-0.5p,-p
Q点坐标有4个:
(-0.5p,0.5p),(-0.5p,p),(-0.5p,-0.5p),(-0.5p,-p)
要∠OQF的最大值,则p|y|/(p^2+2y^2)=最大值
设p|y|/(p^2+2y^2)=s,则
2sy^2-p|y|+sp^2=0
未知数y有实数解,则它的判别式/△≥0,即
(-p)^2-4*2s*sp^2≥0
s^2≤1/8
s最大=1/√8
∠OQF的最大值=arctan(1/√8)