Question

Write the slope of the normal to the curve \[y = \frac{1}{x}\]  at the point \[\left( 3, \frac{1}{3} \right)\] ?

Answer 1

\[\text { Given }: \]

\[y = \frac{1}{x}\]

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

\[\frac{dy}{dx} = \frac{- 1}{x^2}\]

\[\text { Now }, \]

\[\text { Slope of the tangent } = \left( \frac{dy}{dx} \right)_\left( 3, \frac{1}{3} \right) =\frac{- 1}{9}\]

\[\text { Slope of the normal } =\frac{- 1}{\text { Slope of tangent }}=\frac{- 1}{\left( \frac{- 1}{9} \right)}= 9\]