Question

Find the equations of the tangents to the curve y = `- (x + 1)/(x - 1)` which are parallel to the line x + 2y = 6

Answer 1

Curse is y = `(x + 1)/(x - 1)`

DIfferentiating w.r.t. 'x'

`("d"y)/("d"x) = ((x - 1)(1) - (x + 1)(1))/(x - 1)^2`

Slope of the tangent 'm'

= `(x - 1 - x - 1)/(x - 1)^2`

= `- 2/(x - 1)^2`

Given line is x + 2y = 6

Slope of the line = ` 1/2`

Since the tangent is parallel to the line, then the slope of the tangent is `- 1/2`

∴ `("d"y)/("d"x) = 2/(x  1)^2 = - 1/2`

(x – 1)² = 4

x – 1 = ± 2

x = – 1, 3

When x = – 1, y = 0

⇒ point is (– 1, 0)

When x = 3, y = 2

⇒ point is (3, 2)

Equation of tangent with slope `- 1/2` and at the point (– 1, 0) is

 y – 0 = `- 1/2(x + 1)`

2y = – x – 1

⇒ x + 2y + 1 = 0

Equation of tangent with slope ` 1/2` and at the point (3, 2) is 2

y – 2 = `- 1/2 (x - 3)`

2y – 4 = – x + 3

x + 2y – 7 = 0.