Question

If the tangent to the curve $y=x^3+ax+b$ at $(1,-6)$ is parallel to the line $x-y+5=0$, find $a$ and $b$.

Answer 1

Given: $y=x^3+ax+b$ ........(i)

Tangent to the curve= $\frac{dy}{dx}=3x^2+a$

Tangent to the curve at point $(1,-6)$ $=$ ${\frac{dy}{dx}}_{(1,-6)}=3(1{)}^2+a=3+a$

Now, $x-y+5=0$

$\Rightarrow y=x+5$

Comparing it with $y=mx+c$, where $m=$ slope, we get $m=1$

Therefore, slope of the above line is $1$.

Given tangent is parallel to the line $x-y+5=0$.

Thus slope will be equal.

Therefore, $3+a=1$

$\Rightarrow a=1-3=-2$

Substituting $a=-2$ in eqn (i) we get

$-6=1^3+(-2)\left(1\right)+b$

$\Rightarrow -6=1-2+b$

$\Rightarrow -6+1=b$

$\Rightarrow b=-5$

$\Rightarrow a=-2,b=-5$