Question

求曲线｛y=4t² x=1+ 2t-t²在点(1,16)的切线方程和法线方程

Answer 1

1+2t-t²＝1→t＝0或2;

4t²＝16→t＝2或-2.

∴t＝2.

k＝dy/dx＝8t/(2-2t)＝4t/(1-t)＝4×2/(1-2)＝-8.

故切线为y-16＝-8(x-1)→8x+y-24＝0;

法线方程y-16＝(1/8)(x-1)→x-8y+127＝0.