NIMCET 2024 Paper

Let any point on the curve $\mathit{x}\mathit{y}=1$ is $\left(\mathit{t},\frac{1}{\mathit{t}}\right)$, to calculate the slope of tangent at this point, let us differentiate the function,
$\mathit{x}{\mathit{y}}^{\prime }+\mathit{y}=0\Rightarrow {\mathit{y}}^{\prime }=-\frac{\mathit{y}}{\mathit{x}}$
Slope of the tangent $=-\frac{1∕\mathit{t}}{\mathit{t}}=-\frac{1}{{\mathit{t}}^2}$
Slope of the normal $={\mathit{t}}^2$
As ${\mathit{t}}^2>0$, slope of the normal should be positive.