f(x)=ax^2
f'(x)=2ax
依题意f'(1)=2a=2 解得a=1
抛物线方程f(x)=x^2
(1,-3)不在抛物线上
切点在抛物线上,设切点为(x0,x0^2)
(x0^2-(-3))/(x0-1)=2x0
x0^2+3=2x0^2-2x0
x0^2-2x0-3=0
(x0+1)(x0-3)=0
解得两切点为(-1,1) (3,9)
点斜式
(y-1)/(x-(-1))=-2
y=-2x-1
(y-9)/(x-3)=6
y=6x-9
即切线方程为y=-2x-1或y=6x-9