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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 ? - Mathematics

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प्रश्न

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 ?

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उत्तर

\[\text { Given:} \]

\[xy + ax + by = 2 . . . \left( 1 \right)\]

\[\text { On differentiating both sides w.r.t. x, we get }\]

\[x\frac{dy}{dx} + y + a + b\frac{dy}{dx} = 0\]

\[ \Rightarrow \frac{dy}{dx}\left( x + b \right) = - a - y\]

\[ \Rightarrow \frac{dy}{dx}=\frac{- a - y}{x + b}\]

\[\text { Now,} \]

\[ \left( \frac{dy}{dx} \right)_\left( 1, 1 \right) = 2\]

\[ \Rightarrow \frac{- a - 1}{1 + b} = 2\]

\[ \Rightarrow - a - 1 = 2 + 2b\]

\[ \Rightarrow - a = 3 + 2b\]

\[ \Rightarrow a = - \left( 3 + 2b \right)\]

\[\text { On substituting } a= - \left( 3 + 2b \right), x=1 \text { and y = 1 in eq. }(1), \text { we get }\]

\[1 - \left( 3 + 2b \right) + b = 2\]

\[ \Rightarrow 1 - 3 - 2b + b = 2\]

\[ \Rightarrow b = - 4\]

\[\text { and }\]

\[a = - \left( 3 + 2b \right) = - \left( 3 - 8 \right) = 5\]

\[ \therefore a = 5 \text { and }b = - 4\]

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अध्याय 16: Tangents and Normals - Exercise 16.1 [पृष्ठ १०]

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आरडी शर्मा Mathematics [English] Class 12
अध्याय 16 Tangents and Normals
Exercise 16.1 | Q 2 | पृष्ठ १०

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