设A(x1,y1) B(x2,y2) N(-1,0)
M(1,0)
直线L y=k(x-1)
y^2=4x 联立得 x^2-(2+4/k)x+1=0 x1x2=1
kAN=y1/(x1+1) kBN=y2/(x2+1)
kAN+kBN=y1/(x1+1)+y2/(x2+1)=(y1x2+y1+y2x1+y2)/(x1x2+x1+x2+1)
y2=k(x2-1)
y1=k(x1-1)
y1x2+y1+y2x1+y2=k(x1-1)(x2+1)+k(x2-1)(x1+1)=2kx1x2-2k=2k(x1x2-1)=0
kAN+kBN=0
直线AN和直线BN关于x轴对称,x轴是角ANB的角平分线,所以角ANM=角BNM
2.A(x1,y1) M(a,0) AM中点((x1+a)/2,y1/2) 圆心
|AM|=√[(x1-a)^2+y1^2] 直径,半径r=√[(x1-a)^2+y1^2] /2
圆心到直线x=m的距离d=|(x1+a)/2-m|
半弦长=√(r^2-d^2)=√[(a+1-m)x1+m^2-ma]为定值,则a+1-m=0 m=a+1
否存在直线x=m=a+1,使得被以AM为直径的园所截得的弦长为定值