# What is the equation of the line normal to f(x)= -1/(5-x^2)  at x=-3?

Jun 18, 2016

Equation of line normal is $32 x + 12 y + 93 = 0$

#### Explanation:

Slope of the tangent of a curve, at a given point, is given by the value of the first derivative at that point. As normal is perpendicular to tangent, if $m$ is the slope of tangent, slope of normal would be $- \frac{1}{m}$.

To find the point, let us put $x = - 3$ in $f \left(x\right) = - \frac{1}{5 - {x}^{2}}$ and $f \left(- 3\right) = - \frac{1}{5 - {\left(- 3\right)}^{2}} = - \frac{1}{-} 4 = \frac{1}{4}$.

Hence, we are seeking normal at $\left(- 3 , \frac{1}{4}\right)$

$\frac{\mathrm{df}}{\mathrm{dx}} = - \left(- \frac{1}{5 - {x}^{2}} ^ 2\right) \times \left(- 2 x\right) = - \frac{2 x}{5 - {x}^{2}} ^ 2$

Hence slope of tangent at $x = - 3$ is

$- \frac{2 \left(- 3\right)}{5 - {\left(- 3\right)}^{2}} ^ 2 = \frac{6}{- 4} ^ 2 = \frac{6}{16} = \frac{3}{8}$

and slope of normal is $- \frac{1}{\frac{3}{8}} = - \frac{8}{3}$

and equation of line normal is

$\left(y - \frac{1}{4}\right) = - \frac{8}{3} \left(x - \left(- 3\right)\right)$

or $12 y - 3 = - 32 x - 96$ or $32 x + 12 y + 93 = 0$