Question

Find the slope of the normal at the point 't' on the curve \[x = \frac{1}{t}, y = t\] ?

Answer 1

\[\text { Here, } \]

\[x = \frac{1}{t} \text { and } y = t\]

\[\frac{dx}{dt} = \frac{- 1}{t^2}\text { and } \frac{dy}{dt} = 1\]

\[ \therefore \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{1}{\left( \frac{- 1}{t^2} \right)} = - t^2 \]

\[\text { Now }, \]

\[\text { Slope of the tangent }= \left( \frac{dy}{dx} \right)_{} = - t^2 \]

\[\text { Slope of the normal }=\frac{- 1}{\text { Slope of the tangent }}=\frac{- 1}{- t^2}=\frac{1}{t^2}\]