Question

直线L：X=1+tcosA,Y=tsinA,(t为参数）（A不等于0）交抛物线Y^2=8X于M,N两点,求MN中点的轨迹

Answer 1

轨迹方程是直x=-3

X=1+tcosA,Y=tsinA cosA=(X-1)/t sinA=Y/t

(X-1)^2+Y^2=t^2 将Y^2=8X带入

化简后 X^2+6X+1-t^2=0 M(X1,Y1) N(X2,Y2)

X1+X2=-6 中点X0=(X1+X2)/2=-3

MN中点的轨迹方程为X=-3