Question

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 point lies on the curve?

Answer 1

Let the slope of the curve $y=x^3+ax-b$ be $m_1$
$\therefore \frac{dy}{dx}=3x^2+a(\frac{dy}{dx}{)}_{(1,-5)}=3+a=m_1$
Slope of line, $-x+y+4=0$ is $m_2=1$
Since, the given line is perpendicular to the tangent to the given curve at $(1,-5)$
$\therefore m_1\times m_2=-1\Rightarrow (3+a)\left(1\right)=-1\Rightarrow a=-4$
Also, $(1,-5)$ lies on graph.
$\therefore -5=(1{)}^3+a(1)-b\Rightarrow -6=-4-b\Rightarrow b=2$
$\therefore y=x^3-4x-2=0$
$\therefore (2,-2)$ lies on $y=x^3-4x-2=0$