How to find the equation of the normal line to a curve at a given point

Question

How can I find the equation of the normal to the curve y=(5x²)/(1+x²) at the point (2,4)?

Answer 1

normal line to y=5x^2/(1+x^2) at (2,4) has equation

y= -5x/4 + 13/2

= -1.25x + 6.5

slope of the given function is y'= 10x/(1+x^2)^2

at (2,4) slope = 10(2)/(1+2^2)^2 = 20/25 = 4/5

slope of normal line = negative inverse = -5/4 =-1.25

equation of normal line in point slope form is

y-4 = -1.25(x-2)

simplify and rewrite in slope intercept form

y = -1.25x +2.5+4

y = -1.25x +6.5