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Question
Find the slope of the tangent and the normal to the following curve at the indicted point xy = 6 at (1, 6) ?
Solution
\[ xy = 6\]
\[\text { On differentiating both sides w.r.t. x, we get }\]
\[x\frac{dy}{dx} + y = 0\]
\[ \Rightarrow x\frac{dy}{dx} = - y\]
\[ \Rightarrow \frac{dy}{dx} = \frac{- y}{x}\]
\[\text { Now,} \]
\[\text { Slope of the tangent }= \left( \frac{dy}{dx} \right)_\left( 1, 6 \right) =\frac{- y}{x}=\frac{- 6}{1}=-6\]
\[\text{ Slope of the normal }=\frac{- 1}{\left( \frac{dy}{dx} \right)_\left( 1, 6 \right)}=\frac{- 1}{- 6}=\frac{1}{6}\]
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