Question

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

Answer 1

$$\begin{gathered} \mathrm{Here}, \\ x=\frac{1}{t}\mathrm{and}y=t \\ \frac{dx}{dt}=\frac{-1}{t^2}\mathrm{and}\frac{dy}{dt}=1 \\ \therefore \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{1}{\left(\frac{-1}{t^2}\right)}=-t^2 \\ \mathrm{Now}, \\ \text{Slope of the tangent=}{\left(\frac{dy}{dx}\right)}_{}\text{=}-t^2 \\ \text{Slope of the normal=}\frac{-1}{\text{Slope of the tangent}}\text{=}\frac{-1}{-t^2}\text{=}\frac{1}{t^2} \end{gathered}$$