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

Question

10096 views around the world

You can reuse this answer
Creative Commons License

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#.