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

\[\text { Given }:\]

\[x - y + 5 = 0\]

\[ \Rightarrow y = x + 5\]

\[ \Rightarrow \frac{dy}{dx} = 1\]

\[\text { Now,} \]

\[y = x^3 + ax + b . . . \left( 1 \right)\]

\[ \Rightarrow \frac{dy}{dx} = 3 x^2 + a\]

\[\text { Slope of the tangent at }\left( 1, - 6 \right)= \text { Slope of the given line }\]

\[ \Rightarrow \left( \frac{dy}{dx} \right)_\left( 1, - 6 \right) = 1\]

\[ \Rightarrow 3 + a = 1\]

\[ \Rightarrow a = - 2\]

\[\text { On substituting }a= - 2, x=1 \text { and }y=-6 \text { in eq.} (1), \text { we get} \]

\[ - 6 = 1 - 2 + b\]

\[ \Rightarrow b = - 5\]

\[ \therefore a = - 2 \text { and} \ b = - 5\]