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

Question:

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

$x^{2}+3 y+y^{2}=5$ at $(1,1)$

Solution:

Given:

$x^{2}+3 y+y^{2}=5$ at $(1,1)$

Here we have to differentiate the above equation with respect to $x$.

$\Rightarrow \frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}^{2}+3 \mathrm{y}+\mathrm{y}^{2}\right)=\frac{\mathrm{d}}{\mathrm{dx}}(5)$

$\Rightarrow \frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}^{2}\right)+\frac{\mathrm{d}}{\mathrm{dx}}(3 \mathrm{y})+\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{y}^{2}\right)=\frac{\mathrm{d}}{\mathrm{dx}}(5)$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}}\left(\mathrm{x}^{\mathrm{n}}\right)=\mathrm{n} \cdot \mathrm{x}^{\mathrm{n}-1}$

$\Rightarrow 2 x+3 x \frac{d y}{d x}+2 y \times \frac{d y}{d x}=0$

$\Rightarrow 2 x+\frac{d y}{d x}(3+2 y)=0$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}(3+2 \mathrm{y})=-2 \mathrm{x}$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{-2 \mathrm{x}}{(3+2 \mathrm{y})}$

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

$\Rightarrow \frac{d y}{d x}=\frac{-2 \times 1}{(3+2 \times 1)}$

$\Rightarrow \frac{d y}{d x}=\frac{-2}{(3+2)}$

$\Rightarrow \frac{d y}{d x}=\frac{-2}{5}$

$\therefore$ The Slope of the tangent at $(1,1)$ is $\frac{-2}{5}$

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

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

$\Rightarrow$ The Slope of the normal $=\frac{\frac{-1}{-2}}{\frac{-2}{5}}$

$\Rightarrow$ The Slope of the normal $=\frac{5}{2}$