Question

How do you find the equations of the tangent lines for the curve y=(x-1)/(x+1) that are parallel to the line x-2y=3?

Answer 1

The tangents are `x-2y+7=0` and `x-2y-1=0`

Explanation:

As the tangent lines for the given curve are parallel to line `x-2y=3`, which in slope intercept form can be written as `y=1/2x-3/2` and hence has a slope of `1/2`,

the slope of tangents should also be `1/2`.

But slope is given by `f'(x)` the first derivative. As first derivative of `f(x)=(x-1)/(x+1)` is

`f'(x)=(1xx(x+1)-1xx(x-1))/(x+1)^2=2/(x+1)^2`

Hence if `2/(x+1)^2=1/2`, `(x+1)^2=4` or `x+1=+-2`

and values of `x` are `-3` and `1`

For these values of `f(x)` are `(-3-1)/(-3+1)=-4/-2=2` and `(1-1)/(1+1)=0`.

Hence desired tangents are at `(-3,2)` and `(1,0)`

and these are `(y-2)=1/2(x+3)` or `2y-4=x+3` i.e. `x-2y+7=0`

and `(y-0)=1/2(x-1)` or `2y=x-1` i.e. `x-2y-1=0`
graph{((x-1)/(x+1)-y)(x-2y-3)(x-2y+7)(x-2y-1)=0 [-10.96, 9.04, -4.12, 5.88]}