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Question
Find the values of a and b if the slope of the tangent to the curve xy + ax + by = 2 at (1, 1) is 2 ?
Solution
\[\text { Given:} \]
\[xy + ax + by = 2 . . . \left( 1 \right)\]
\[\text { On differentiating both sides w.r.t. x, we get }\]
\[x\frac{dy}{dx} + y + a + b\frac{dy}{dx} = 0\]
\[ \Rightarrow \frac{dy}{dx}\left( x + b \right) = - a - y\]
\[ \Rightarrow \frac{dy}{dx}=\frac{- a - y}{x + b}\]
\[\text { Now,} \]
\[ \left( \frac{dy}{dx} \right)_\left( 1, 1 \right) = 2\]
\[ \Rightarrow \frac{- a - 1}{1 + b} = 2\]
\[ \Rightarrow - a - 1 = 2 + 2b\]
\[ \Rightarrow - a = 3 + 2b\]
\[ \Rightarrow a = - \left( 3 + 2b \right)\]
\[\text { On substituting } a= - \left( 3 + 2b \right), x=1 \text { and y = 1 in eq. }(1), \text { we get }\]
\[1 - \left( 3 + 2b \right) + b = 2\]
\[ \Rightarrow 1 - 3 - 2b + b = 2\]
\[ \Rightarrow b = - 4\]
\[\text { and }\]
\[a = - \left( 3 + 2b \right) = - \left( 3 - 8 \right) = 5\]
\[ \therefore a = 5 \text { and }b = - 4\]
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