# How do you give the equation of the normal line to the graph of y=2xsqrt(x^2+8)+2 at point (0,2)?

Mar 1, 2015

The answer is: $y = - \frac{\sqrt{2}}{8} x + 2$.

The normal line to the graph in one point is the perpendicular at the tangent line to the graph in that point.

Remembering that two lines $n$ and $t$ are perpendicular if and only if this rule is verified:

${m}_{n} = - \frac{1}{m} _ t$ , where $m$ is the slope,

and

the slope of the tangent line in a point to a curve is the first derivative in that point,

$y ' = 2 \left(1 \cdot \sqrt{{x}^{2} + 8} + x \cdot \frac{1}{\sqrt{{x}^{2} + 8}} \cdot 2 x\right)$

and so:

$y ' \left(0\right) = 2 \sqrt{8} = 4 \sqrt{2}$.

So:

${m}_{t} = 4 \sqrt{2} \Rightarrow {m}_{n} = - \frac{1}{4 \sqrt{2}} = - \frac{1}{4 \sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = - \frac{\sqrt{2}}{8}$.

The line that passes from a given a point $P \left({x}_{P} , {y}_{P}\right)$ and with a slope $m$, is:

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

so:

$y - 2 = - \frac{\sqrt{2}}{8} \left(x - 0\right) \Rightarrow y = - \frac{\sqrt{2}}{8} x + 2$.