设过A点直线方程为y=Kx-aK,
与抛物线相切,即x^2+1=Kx-aK
化为x^2-Kx+1+aK=0只有一解
则△=0,即K^2-4(1+aK)=0K^2+4aK-4=0
解得K1=[-4a+√(16a^2+16)]/2=-2a+2√(a^2+1)
K2=-2a-2√(a^2+1)
K1*K2=-4问题1得证
K1+K2=-4a
x=K/2,将K1k2代入得x1=K1/2=-a+√(a^2+1)x2=K2/2=-a-√(a^2+1)
则x1+x2=(K1+K2)/2=-2ax1*x2=-1
设切点为P(x1,y1),Q(x2y2)
PQ方程为y=[(y2-y1)/(x2-x1)]*x+y1-[(y2-y1)/(x2-x1)]*x1
y=[(x2^2-x1^2)/(x2-x1)]*x+x1^2+1-[(x2^2-x1^2)/(x2-x1)]*x1
y=(x2+x1)*x+x1^2+1-(x2+x1)*x1
y=(x2+x1)*x+x1^2+1-(x2+x1)*x1
y=-2a*x+x1^2+1-x1*x2-x1^2
y=-2a*x+1-x1*x2
y=-2a*x+2
此方程即PQ过定点(0,2)问题2得证
SAPQ=(y1+y2)(x1-x2)/2-(x1-a)*y1/2-(a-x2)*y2/2
时间不够,下次再解