Given:

The tangent to the curve

  y=x−1x+1

Parallel to the line

  x−2y=2

Concept used:

The equation is in slope −intercept form, y=mx+c

An equation for the line through the point (x1,y1) with slope m is

  y−y1=m(x−x1)

Calculation:

The function

  y=x−1x+1.......................(1)

The derivative of a function

  y=f(x)y′=f′(x)=dydx=m

Differentiating the equation (1) with respect to x

  y=x−1x+1y′=(x+1)ddx(x−1)−(x−1)ddx(x+1)(x+1)2y′=(x+1)−(x−1)(x+1)2y′=2(x+1)2........................(2)

The derivative is slope of the tangent line so in order to the slope of the tangent line

The derivative of constant is zero

  x−2y=2y=−2+x2y=−1+x2y′=12.............................(3)

The equation is in slope −intercept form

The slope of the line is =12

From equation (2) and equation (3)

  12=2(x+1)24=(x2+2x+1)x2+2x−3=0x2+3x−x−3=0x(x+3)−1(x+3)=0(x−1)(x+3)=0x=1,−3

An equation for the line through the point (x1,y1) with slope m is

  y−y1=m(x−x1)y−1=12(x+3)  y=12(x+3)+1