Find the slope of the normal


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


Given that the curve $x=\frac{1}{t}, y=t$

$\frac{\mathrm{dx}}{\mathrm{dt}}=\frac{-1}{\mathrm{t}^{2}}, \frac{\mathrm{dy}}{\mathrm{dt}}=1$

$\therefore \frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}=\frac{1}{\frac{-1}{t^{2}}}=-t^{2}$

Now, Slope of tangent $=-t^{2}$

Slope of normal $=\frac{-1}{\text { Slope of tangent }}=\frac{1}{t^{2}}$

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