已知抛物线x2=2py(p>0),过焦点F的动直线l交抛物线于AB两点,抛物线在AB两点处的切线相交于点Q
1个回答

∵F(0,p/2),设直线方程为y=kx+(p/2),

将x^2=2py与y=kx+(p/2)联立,解得:x^2 - 2pkx -(p^2)=0

设A(x1,y1)B(x2,y2)则:x1+x2= 2px ,x1x2= - p^2

所以 y1y2= k(x^2) + (p/2)=(k^2)x1x2+(kp/2)(x1+x2)+(p^2)/4= - (k^2)(p^2) + (k^2)(p^2) + (p^2)/4 =(p^2)/4

所以向量OA * 向量OB=x1x2+y1y2= - (3/4) / (p^2)