If the tangent to the curve, $y=x^3+ax-b$ at the point $(1, -5)$ is perpendicular to the line, $-x+y+4=0$, then which one of the following points lies on the curve?

$y=x^2+ax-b$

$(1, -5)$ lies on the curve

$\Rightarrow -5=1+a-b\Rightarrow a-b=-6$ .(i)

Also, $y'=3x^2+a$

$y'_{(1, -5)}=3+a$ (slope of tangent)

$\because$ this tangent is $\perp$ to $-x+y+4=0$

$\Rightarrow (3+a)(1)=-1$

$\Rightarrow a=-4$ .(ii)

By (i) and (ii): $a=-4$, $b=2$

$\therefore y=x^3-4x-2$

$(2, -2)$ lies on this curve.