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

Oct 22, 2017

$y = \sqrt{2} x - \sqrt{2}$

#### Explanation:

I always prefer to rewrite my functions so I can simply use the Product Rule.

$f \left(x\right) = x {\left({x}^{2} - 2\right)}^{-} \left(\frac{1}{2}\right)$

Find the derivative to find the slope of the tangent line at $x = 2$.

$f ' \left(x\right) = {\left({x}^{2} - 2\right)}^{-} \left(\frac{1}{2}\right) - \frac{x}{2} {\left({x}^{2} - 2\right)}^{- \frac{3}{2}} \left(2 x\right)$

$f ' \left(x\right) = \frac{1}{\sqrt{{x}^{2} - 2}} - {x}^{2} / {\left({x}^{2} - 2\right)}^{\frac{3}{2}}$

$f ' \left(2\right) = \frac{1}{\sqrt{{2}^{2} - 2}} - {2}^{2} / {\left({2}^{2} - 2\right)}^{\frac{3}{2}}$

$f ' \left(2\right) = \frac{1}{\sqrt{2}} - \frac{4}{2} ^ \left(\frac{3}{2}\right) = - \frac{1}{\sqrt{2}} = {m}_{t}$

To find the normal slope, we take the negative reciprocal of our tangent slope.

${m}_{n} = \sqrt{2}$

We can now construct an equation using a basic point-slope formula:

$y - {y}_{o} = m \left(x - {x}_{o}\right)$

${x}_{o}$ is given to us: ${x}_{o} = 2$

${y}_{o}$ can be solved by plugging in ${x}_{o}$ back into our original equation:

${y}_{o} = f \left(2\right) = \left(2\right) {\left({2}^{2} - 2\right)}^{-} \left(\frac{1}{2}\right) = \frac{2}{\sqrt{2}}$

$y - \frac{2}{\sqrt{2}} = \sqrt{2} \left(x - 2\right)$

$y = \sqrt{2} x - 2 \sqrt{2} + \frac{2}{\sqrt{2}}$

$y = \sqrt{2} x - \sqrt{2}$