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