Question

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

Answer 1

Equation of normal at `x=-1` is `27x+6y+29=0`

Explanation:

As `f(x)=x^2/(1+4x)`, at `x=-1`, we have `f(-1)=(-1)^2/(1+4(-1))=1/-3=-1/3`

And hence normal passes through `(-1,-1/3)`

Now as `f(x)=x^2/(1+4x)`,

`(df)/(dx)=((1+4x)xx2x-4xxx^2)/(1+4x)^2=((2x+8x^2)-4x^2)/(1+4x)^2`

`(df)/(dx)=(2x(1+2x))/(1+4x)^2`

and hence slope of curve i.e. tangent at `x=-1` is

`(2(-1)(1+2(-1)))/(1+4(-1))^2` or `((-2)xx(-1))/(-3)^2=2/9`

and slope of normal would be `-1/(2/9)=-9/2`

Hence, equation of normal at `x=-1` is given by `(y+1/3)=-9/2xx(x+1)`

or `6(y+1/3)=-9xx6/2xx(x+1)` or `6y+2=-27x-27` or

`27x+6y+29=0`