Question

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

Answer 1

`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`