Question

抛物线y^2=2px(p>0),设AB为焦点弦，M为准线与x轴的交点，F为焦点,求证∠AMF=∠BMF.

如题.
请详解。

Answer 1

y^2=2px(p>0),设AB为焦点弦，M为准线与x轴的交点，F为焦点
F(0.5p,0),M(-0.5p,0)
A(2pa^2,2pa),B(2pb^2,2pb)
k(AB)=(2pa-2pb)/(2pa^2-2pb^2)=1/(a+b)
k(AF)=2pa/(2pa^2-0.5p)=4a/(4a^2-1)
k(AB)=k(AF)
1/(a+b)=4a/(4a^2-1)
4ab=-1
b=-1/(4a)
4b=-1/a,4b^2=1/(4a^2),4b^2+1=(1+4a^2)/(4a^2)
k(AM)=2pa/(2pa^2+0.5p)=4a/(4a^2+1)
k(BM)=2pb/(2pb^2+0.5p)=4b/(4b^2+1)=(-1/a)/[(1+4a^2)/(4a^2)]=-4a/(4a^2+1)
∵k(AM)=-k(BM)
∴∠AMF=∠BMF

Answer 2

见图