How many tangent lines to the curve #y = x/(x+1)# pass through the point (1,2)?

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Answer 1 · 2018-04-07T18:40:47

There are 2 tangent lines that pass through the point #(1,2)#.

#y = 1/(-1+sqrt3)^2(x-1)+2#

and

#y = 1/(-1-sqrt3)^2(x-1)+2#

Given: #y = x/(x+1)#

The point-slope form of the equation of a line tells us that the form of the tangent lines must be:

#y = m(x-1)+2" [1]"#

For the lines to be tangent to the curve, we must substitute the first derivative of the curve for #m#:

#dy/dx = ((d(x))/dx(x+1)-x(d(x+1))/dx)/(x+1)^2#

#dy/dx = (x+1-x)/(x+1)^2#

#dy/dx = 1/(x+1)^2#

#m = 1/(x+1)^2" [2]"#

Substitute equation [2] into equation [1]:

#y = (x-1)/(x+1)^2+2" [1.1]"#

Because the line must touch the curve, we may substitute #y = x/(x+1)#:

#x/(x+1) = (x-1)/(x+1)^2+2#

Solve for x:

#x(x+1) = (x-1)+2(x+1)^2#

#x^2+x = x -1 +2x^2+4x+2#

#x^2+4x+1#

#x = (-4+-sqrt(4^2-4(1)(1)))/(2(1))#

#x = -2+-sqrt(3)#

#x = -2+sqrt(3)# and #x = -2-sqrt(3)#

There are 2 tangent lines.

#y = 1/(-1+sqrt3)^2(x-1)+2#

and

#y = 1/(-1-sqrt3)^2(x-1)+2#