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

1 Answer
Mar 1, 2015

The answer is: y=-sqrt2/8x+2y=28x+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 nn and tt are perpendicular if and only if this rule is verified:

m_n=-1/m_tmn=1mt , where mm 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.