# What is the equation of the normal line of f(x)=x/(2-x^2) at x = 3?

May 2, 2018

$77 y = - 343 x + 996$

#### Explanation:

First we find the y-coordinates;
Substitute x=3
$\frac{3}{2 - {\left(3\right)}^{2}} = - \frac{3}{7}$

Then differentiate $\frac{x}{2 - {x}^{2}}$
$\frac{\mathrm{dy}}{\mathrm{dx}}$=$\frac{{x}^{2} + 2}{{x}^{2} - 2} ^ 2$

Substitute 3 to find the gradient of the tangent
$\frac{\mathrm{dy}}{\mathrm{dx}}$=$\frac{{3}^{2} + 2}{{3}^{2} - 2} ^ 2$
=$\frac{11}{49}$

General equation of normal:
$\left(y - {y}_{1}\right)$=-1/(gradient of tangent)$\left(x - {x}_{1}\right)$

Substituting the values
$\left(y - \left(- \frac{3}{7}\right)\right) = - \frac{1}{\frac{11}{49}} \left(x - 3\right)$
$y + \frac{3}{7} = - \frac{49}{11} x + \frac{147}{11}$
$y = - \frac{49}{11} x + \frac{996}{77}$
$77 y = - 343 x + 996$