y^2－4ty－4=0,点F﹙2t^2﹢1,2t﹚,点N﹙[2／t^2]﹢1,﹣2／t﹚,由两点式得FN的方程为x﹣﹙t﹣1／t﹚y＝3方程要怎么算出来?

Question

y^2－4ty－4=0,点F﹙2t^2﹢1,2t﹚,点N﹙[2／t^2]﹢1,﹣2／t﹚,由两点式得FN的方程为x﹣﹙t﹣1／t﹚y＝3
方程要怎么算出来?

Answer 1

先算FN的斜率 ,K=(2t+2／t)/[(2t^2﹢1)-(2／t^2﹢1)]=(t+1/t0/(t^2-1/t^2)=1/(t-1/t)
再代入直线方程的点斜式,得
y-2t=1/(t-1/t)*[x-(2t^2+1)]
(t-1/t)y-2t(t-1/t)=x-(2t^2+1)
x-(t-1/t)y=(2t^2+1)-2t(t-1/t)=2t^2+1-2t^2+2=3
即x-(t-1/t)y=3