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प्रश्न
Find the slope of the normal at the point 't' on the curve \[x = \frac{1}{t}, y = t\] ?
उत्तर
\[\text { Here, } \]
\[x = \frac{1}{t} \text { and } y = t\]
\[\frac{dx}{dt} = \frac{- 1}{t^2}\text { and } \frac{dy}{dt} = 1\]
\[ \therefore \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{1}{\left( \frac{- 1}{t^2} \right)} = - t^2 \]
\[\text { Now }, \]
\[\text { Slope of the tangent }= \left( \frac{dy}{dx} \right)_{} = - t^2 \]
\[\text { Slope of the normal }=\frac{- 1}{\text { Slope of the tangent }}=\frac{- 1}{- t^2}=\frac{1}{t^2}\]
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