Find equations of the tangent lines to the curve y = (x - 1) / (x + 1) that are parallel to the line x - 2y = 2

Solution:

Given, the equation of the line is x - 2y = 2

Converting the equation in slope intercept form,i.e. y = mx + c

x - 2 = 2y

y = (1/2)x - 1

Here, slope m = 1/2

Slope of the tangent line must be 1/2.

Now find derivative,

dy/dx = d(x - 1) / (x + 1) /dx

dy/dx = [(x + 1) (1) - (x - 1) (1)] / (x + 1)²

On simplification,

dy/dx = [x + 1 - x + 1] / (x + 1)²

dy/dx = 2 / (x + 1)²

We know dy/dx i.e. slope = 1/2

1/2 = 2 / (x + 1)²

(x + 1)² = 4

Taking square root,

√(x + 1)² = √4

x + 1 = ±2

x + 1 = +2

x = 1

x + 1 = -2

x = -3

When x = 1, 

y = (1 - 1) / (1 + 1)

y = 0

When x = -3,

y = (-3 - 1) / (-3 + 1)

y = -4/-2

y = 2

Equation of the line can be found by (y - y₁) = m(x - x₁)

At (1,0) and m=1/2

(y - 0) = (1/2)(x - 1)

y = x/2 - 1/2

At (-3,2) and m = 1/2

(y - 2) = (1/2)(x - (-3))

y - 2 = (1/2) (x + 3)

y - 2 = x/2 + 3/2

y = x/2 + 3/2 + 2

y = x/2 + 7/2

Therefore, the equations of the tangent line are y = x/2 - 1/2 and y = x/2 + 7/2.

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Find equations of the tangent lines to the curve y = (x - 1) / (x + 1) that are parallel to the line x - 2y = 2

Summary:

The equations of the tangent lines to the curve y=(x - 1) / (x + 1) that are parallel to the line x - 2y = 2 are y = x/2 - 1/2 and y = x/2 + 7/2.