Question

Find the equations of the tangents to the curve y = ${\displaystyle -\frac{x+1}{x-1}}$ which are parallel to the line x + 2y = 6

Answer 1

Curse is y = ${\displaystyle \frac{x+1}{x-1}}$

DIfferentiating w.r.t. 'x'

${\displaystyle \frac{\text{d}y}{\text{d}x}=\frac{(x-1)\left(1\right)-(x+1)\left(1\right)}{{(x-1)}^2}}$

Slope of the tangent 'm'

= ${\displaystyle \frac{x-1-x-1}{{(x-1)}^2}}$

= ${\displaystyle -\frac{2}{{(x-1)}^2}}$

Given line is x + 2y = 6

Slope of the line = ${\displaystyle \frac{1}{2}}$

Since the tangent is parallel to the line, then the slope of the tangent is ${\displaystyle -\frac{1}{2}}$

∴ ${\displaystyle \frac{\text{d}y}{\text{d}x}=\frac{2}{{(x 1)}^2}=-\frac{1}{2}}$

(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 ${\displaystyle -\frac{1}{2}}$ and at the point (– 1, 0) is

 y – 0 = ${\displaystyle -\frac{1}{2}(x+1)}$

2y = – x – 1

⇒ x + 2y + 1 = 0

Equation of tangent with slope ${\displaystyle \frac{1}{2}}$ and at the point (3, 2) is 2

y – 2 = ${\displaystyle -\frac{1}{2}(x-3)}$

2y – 4 = – x + 3

x + 2y – 7 = 0.