巧用抛物线的一个性质解决一类高考题

定理从抛物线外一点P引抛物线的两条切线PA,PB,切点分别为A,B,若F是抛物线的焦点,则有∠PFA=∠PFB.图1证法1如图1,设抛物线的方程为x2=2py(p>0),点P(x0,y0),A(x1,y1),B(x2,y2),则切线PA:x1x-py-py1=0,∴x1x0-py0-py1=0.切线PB:x2x-py-py2=0,∴x2x0-py0-py2=0.∵FA=(x1,y1-p2),FB=(x2,y2-p2),FP=(x0,y0-p2),∴cos∠PFA=FA·FP|FA||·FP|=x1x0 (y1-p2)(y0-p2)|FP|(y1 p2)=p(y1 y0) (y1-p2)(y0-p2)|FP|(y1 p2)=y1y0 p2(y1 y0) 2p4|FP|(y1 p2)=(y1 p2)(y0 p2)|FP|(y1 p2)=y0 p2|FP|,同理cos∠PFB=y0 p2|FP|,∴∠PFA=∠P…