Find the The Slopes of the tangent and the normal to the following curves at the indicated points :

Solution:

Given:

$x y=56$ at $(1,6)$

Here we have to use the product rule for above equation.

If $u$ and $v$ are differentiable function, then

$\frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{UV})=\mathrm{U} \times \frac{\mathrm{dV}}{\mathrm{dx}}+\mathrm{V} \times \frac{\mathrm{dU}}{\mathrm{dx}}$

$\frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{xy})=\frac{\mathrm{d}}{\mathrm{dx}}(6)$

$\Rightarrow \times \times \frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{y})+\mathrm{y} \times \frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{x})=\frac{\mathrm{d}}{\mathrm{dx}}(5)$

$\therefore \frac{\mathrm{d}}{\mathrm{dx}}($ Constant $)=0$

$\Rightarrow \mathrm{x} \frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{y}=0$

$\Rightarrow \mathrm{x} \frac{\mathrm{dy}}{\mathrm{dx}}=-\mathrm{y}$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{-\mathrm{y}}{\mathrm{x}}$

The Slope of the tangent at $(1,6)$ is

$\Rightarrow \frac{d y}{d x}=\frac{-6}{1}$

$\Rightarrow \frac{d y}{d x}=-6$

$\therefore$ The Slope of the tangent at $(1,6)$ is $-6$

$\Rightarrow$ The Slope of the normal $=\frac{-1}{\text { The Slope of the tangent }}$

$\Rightarrow$ The Slope of the normal $=\frac{-1}{\left(\frac{\mathrm{d} y}{\mathrm{dx}}\right)}$

$\Rightarrow$ The Slope of the normal $=\frac{-1}{-6}$

$\Rightarrow$ The Slope of the normal $=\frac{1}{6}$