Find equations of the tangent lines to the curve y = x - 1/ x + 1 that are parallel to the line x - 2y = 5.

Solution:

Given, y = (x - 1)/(x + 1)

Slope of tangent can be found by functional derivative i.e. dy/dx

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

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

dy/dx = [x + 1 - x + 1]/(x + 1)²

dy/dx = 2/(x + 1)² --- (1)

The slope intercept form of a linear equation is given by y = mx + c

Where m = slope and c = constant

Slope of tangent parallel to the line is given by 1/m

Tangent is parallel to line x - 2y = 5.

Grouping above equation into slope intercept form, we get

⇒ x - 5 = 2y

Dividing both sides by 2, we get

⇒ y = 1/2 x - 5/2

Slope of tangent parallel to the line x - 2y = 5 is found as

⇒ 1/m = 1/2 --- (2)

From (1) and (2), we get

⇒ 2/(x + 1)² = 1/2

By simplification using cross multiplication,

⇒ (x + 1)² = 4

Taking square root, we get

⇒ √(x + 1)² = √4

⇒ x + 1 = ±2

When x + 1 = 2, we get x = 1

When x + 1 = -2, we get x = -3

This implies that we will have two tangents parallel to x - 2y = 5.

Substituting the values of x in the equation y = x - 1/ x+ 1

At x = 1, y = (1 - 1)/(1 + 1)

y = 0

Tangent to curve is at (1, 0)

At x = -3, y = (-3 - 1)/(-3 + 1)

y = -4/-2 = 2

Tangent to curve is at (-3,2)

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

At (1, 0) equation of tangent line is (y - 0) = 1/2(x - 1)

On simplification, 2y = x - 1

Further at x = -3, equation of tangent line is (y - 2) = 1/2(x + 3)

On simplification, 2(y - 2) = x + 3

⇒ 2y - 4 = x + 3

⇒ 2y = x + 7

Therefore, two tangent lines are 2y = x - 1 and 2y = x + 7.

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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 = 5.

Summary:

The equations of the tangent lines to the curve y = x - 1/x+ 1 that are parallel to the line x - 2y = 5 are 2y = x - 1 and 2y = x + 7.