How do you find the equations of the tangent lines to the curve #y= (x-1)/(x+1)# that are parallel to the line x-2y=5?

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Answer 1 · 2016-03-26T07:16:55

We have two tangents which are

#x-2y-1=0# and #x-2y+7=0#

As #y=(x-1)/(x+1)# and slope of tangent s given by function's derivative, let us first find #(dy)/(dx)#, which is given by

#((x+1)xx1-1xx(x-1))/(x+1)^2# or #2/(x+1)^2#

As the tangent is parallel to line #x-2y=5#, whose slope is #1/2#

(we get this converting equation of given line to slope intercept form i.e. #y=1/2x-5/2#)

Hence we should have #2/(x+1)^2=1/2# or #(x+1)^2=4#, which gives us #x=1# or #x=-3# i.e. we will have two tangents parallel to #x-2y=5#

Further at #x=1#, #y=(1-1)/(1+1)=0# i.e. tangent to curve is at #(1,0)#

and at #x=-3#, #y=(-3-1)/(-3+1)=-4/-2=2# i.e. curve passes through #(-3,2)#

Hence the two tangents will be #(y-0)=1/2(x-1)# or #2y=x-1#

and #(y-2)=1/2(x+3)# or #2y-4=x+3#

or #x-2y-1=0# and #x-2y+7=0#