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

Nov 15, 2015

$y = \frac{1}{3 {e}^{9}} x - \frac{1}{e} ^ 9 + {e}^{9}$

#### Explanation:

If tangent line is $y = k x + n$ then $k = f ' \left({x}_{0}\right)$ where ${x}_{0} = 3$.

$f ' \left(x\right) = {e}^{{x}^{2} / \left(x - 2\right)} \cdot \frac{2 x \left(x - 2\right) - {x}^{2}}{x - 2} ^ 2$

$k = f ' \left(3\right) = - 3 {e}^{9}$

Equation of the normal is $y - {y}_{0} = {k}_{1} \left(x - {x}_{0}\right)$ where:

${x}_{0} = 3 \implies {y}_{0} = f \left({x}_{0}\right) = f \left(3\right) = {e}^{9}$ and

$k \cdot {k}_{1} = - 1 \implies {k}_{1} = - \frac{1}{k} = \frac{1}{3 {e}^{9}}$

$y - {e}^{9} = \frac{1}{3 {e}^{9}} \left(x - 3\right)$

$y = \frac{1}{3 {e}^{9}} \left(x - 3\right) + {e}^{9}$

$y = \frac{1}{3 {e}^{9}} x - \frac{1}{e} ^ 9 + {e}^{9}$