Question

直线y=-x+1与抛物线y=x^2-a有两个不同交点 抛物线在两点处切线分别为L1,L2.求a值变化时L1与L2交点的轨

Answer 1

y=-x+1与y=x^2-a联立消去y

得x²-a=-x+1

即x²+x-a-1=0

Δ=1+4(a+1)>0

设交点为A(x1,y1),B(x2,y2)

∴x1+x2=-1,x1x2=-(a+1)

对y=x²-a求导得,y'=2x

L1:y-y1=2x1(x-x1) ①

L2:y-y2=2x2(x-x2) ②

①-②:

y2-y1=2(x1-x2)x+2(x²2-x²1)

∵y1=-x1+1,y2=-x2+1

∴x1-x2=2(x1-x2)x+2(x2+x1)(x2-x1)

∵x1≠x2

∴1=2x-2(x1+x2)=2x+2

∴x=-1/2

即L1,L2交点的横坐标为常数-1/2

将 x=-1/2代入y=-x+1,

得y=3/2

∴L1,L2交点的方程为x=-1/2( y