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Question
Write the slope of the normal to the curve \[y = \frac{1}{x}\] at the point \[\left( 3, \frac{1}{3} \right)\] ?
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Solution
\[\text { Given }: \]
\[y = \frac{1}{x}\]
\[\text { On differentiating both sides w.r.t.x, we get }\]
\[\frac{dy}{dx} = \frac{- 1}{x^2}\]
\[\text { Now }, \]
\[\text { Slope of the tangent } = \left( \frac{dy}{dx} \right)_\left( 3, \frac{1}{3} \right) =\frac{- 1}{9}\]
\[\text { Slope of the normal } =\frac{- 1}{\text { Slope of tangent }}=\frac{- 1}{\left( \frac{- 1}{9} \right)}= 9\]
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