Question

过点P(1,0)倾斜角为60°的直线与圆x^2+y^2=4相交于A,B两点,求向...

过点P(1,0)倾斜角为60°的直线与圆x^2+y^2=4相交于A,B两点,求向量PA点乘向量PB

Answer 1

设直线方程为y=kx+m,k=tan60°=√3,P（1,0）代入直线方程,得：m=-√3,直线方程为：y=√3x-√3,设A（x1,y1),B(x2,y2),代入直线方程,x^2+3(x-1)^2=4,4x^2-6x-1=0,根据韦达定理,x1+x2=3/2,x1*x2=-1/4,向量PA=（x1-2,y1),向量PB=(x2-2,y2),PA·PB=x1*x2-2(x1+x2)+4+y1*y2=(-1/4)-2(3/2)+4+(√3x1-√3)(√3x2-√3)=-1/4-3+4+3(-1/4-3/2+1)=-5/4-9/4=-7/2.