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Question
Find the slope of the tangent and the normal to the following curve at the indicted point x2 + 3y + y2 = 5 at (1, 1) ?
Solution
\[x^2 + 3y + y^2 = 5\]
\[\text { On differentiating both sides w.r.t.x, we get }\]
\[2x + 3\frac{dy}{dx} + 2y \frac{dy}{dx} = 0\]
\[ \Rightarrow \frac{dy}{dx}\left( 3 + 2y \right) = - 2x\]
\[ \Rightarrow \frac{dy}{dx} = \frac{- 2x}{3 + 2y}\]
\[\text { Now,} \]
\[\text { Slope of the tangent }= \left( \frac{dy}{dx} \right)_\left( 1, 1 \right) =\frac{- 2x}{3 + 2y}=\frac{- 2}{3 + 2}=\frac{- 2}{5}\]
\[\text { Slope of the normal }=\frac{- 1}{\left( \frac{dy}{dx} \right)_\left( 1, 1 \right)}=\frac{- 1}{\left( \frac{- 2}{5} \right)}=\frac{5}{2}\]
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