A.4x-y-3=0 B.x+4y-5=0
C.4x-y+3=0 D.x+4y+3=0
答案:
则4x03=4,则x03=1,x0=1,y0=14=1.故切点坐标为(1,1).设切线方程为y=4x+b,
则代入切点坐标求得切线方程为4x-y-3=0.