Question

已知直线L经过点P（1,1）,倾斜角α=π／6（1）写出直线L的参数方程（2）设L与圆x^2+y^2=4相交与两点A,B

Answer 1

1.∵直线过点P（1,1）,倾斜角为π/6,

∴直线方程为y-1=tan(π/6)(x-1)

(y-1)/sin(π/6) =(x-1)/cosπ/6

令t=(y-1)/sin(π/6) =(x-1)/cosπ/6

则x=1+t×cosπ/6,y=1+t×sin(π/6)

∴参数方程为x=1+(√3t)/2,y=1+t/2,（t为参数）.2.

x=1+(√3t)/2

把直线 代入x^2+y^2=4

y=1+t/2得 [1+(√3t)/2]^2+(1+t/2)^2=4

t^2+(√3+1)t-2=0 t1t2=-2

则点P到A,B两点的距离之积为2.