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?

Question

Charles Arocha · Accepted Answer

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#

Samantha Adkison · Answer

undefined To find the equations of the tangent lines to the curve y = (x-1)/(x+1) that are parallel to the line x-2y=5, we need to follow these steps:

1. Differentiate the given curve equation y = (x-1)/(x+1) to find its derivative.
2. Set the derivative equal to the slope of the given line x-2y=5, since parallel lines have the same slope.
3. Solve the resulting equation for x to find the x-coordinate(s) of the point(s) where the tangent lines intersect the curve.
4. Substitute the x-coordinate(s) into the original curve equation to find the corresponding y-coordinate(s).
5. Use the point-slope form of a line, y - y1 = m(x - x1), where (x1, y1) is a point on the curve and m is the slope, to write the equations of the tangent lines.

By following these steps, you will obtain the equations of the tangent lines to the curve y = (x-1)/(x+1) that are parallel to the line x-2y=5.