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

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Answer 1 · 2015-03-01T18:52:49

The answer is: $y=-sqrt2/8x+2$ .

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:

$m_n=-1/m_t$ , 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,

$y'=2(1*sqrt(x^2+8)+x*1/sqrt(x^2+8)*2x)$

and so:

$y'(0)=2sqrt8=4sqrt2$ .

So:

$m_t=4sqrt2rArrm_n=-1/(4sqrt2)=-1/(4sqrt2)*sqrt2/sqrt2=-sqrt2/8$ .

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:

$y-2=-sqrt2/8(x-0)rArry=-sqrt2/8x+2$ .

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