How do you give the equation of the normal line to the graph of y=2xx2+8+2 at point (0,2)?

Question

aludirao09au · Accepted Answer

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 (1 \cdot \sqrt{x^{2} + 8} + x \cdot \frac{1}{\sqrt{x^{2} + 8}} \cdot 2 x)

and so:

y^{'} (0) = 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 (x_{p}, y_{p})

and with a slope m, is:

y - y_{p} = m (x - x_{p})

so:

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

.

How do you give the equation of the

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