# What is the equation of the normal line of #f(x)=x^2/(4x-1)# at #x=4#?

##### 1 Answer

# y = 225/8 x -2233/20 #

#### Explanation:

To find the equation of the normal , require to find the gradient of the tangent (m) and (a , b ) a point on the line. Evaluating f'(4) will provide m , and evaluating (f(4) will give (a , b ).

differentiating f(x) using the

#color(blue)(" quotient rule ")# If fx)

# = g(x)/(h(x)) #

then f'(x)# =( h(x).g'(x) - g(x).h'(x))/( (h(x))^2# hence f'(x) =

# ((4x - 1) d/dx(x^2) - x^2 d/dx(4x-1))/(4x - 1 )^2 #

# =( (4x-1).2x - x^2 .4)/(4x-1)^2 =( 8x^2 - 2x-8x^2)/(4x-1)^2 # hence f'(x)

# = -(2x)/(4x-1)^2#

and f'(4) =#-8/225 # now f(4) =

#16/15 rArr (a,b) = (4 , 16/15)#

The product of the gradients of the tangent and the normal equal -1.so m of normal

#xx -8/225 = -1color(black)(" m normal ") = 225/8# equation of normal : y - b = m(x - a)

y -

#16/15 = 225/8(x - 4 )#

#rArr y = 225/8 x - 2233/20 #