Question

If the tangent to the curve y = x³ + ax + b at (1, − 6) is parallel to the line x − y + 5 = 0, find a and b.

Answer 1

$$\begin{gathered} \text{Given:} \\ x-y+5=0 \\ \Rightarrow y=x+5 \\ \Rightarrow \frac{dy}{dx}=1 \\ \mathrm{Now}, \\ y=x^3+ax+b...\left(1\right) \\ \Rightarrow \frac{dy}{dx}=3x^2+a \\ \text{Slope of the tangent at}\left(1,-6\right)\text{= Slope of the given line} \\ \Rightarrow {\left(\frac{dy}{dx}\right)}_{\left(1,-6\right)}=1 \\ \Rightarrow 3+a=1 \\ \Rightarrow a=-2 \\ \text{On substituting}\mathit{\text{a=}}-2,\mathit{\text{x}}\text{=1 and}\mathit{\text{y}}\text{=-6 in eq. (1), we get} \\ -6=1-2+b \\ \Rightarrow b=-5 \\ \therefore a=-2\text{and}b=-5 \end{gathered}$$