Find the slopes of the tangent lines to the curve #y^2 −x +1 = 0# at the points (2,−1) and (2,1)?

1 Answer
Apr 22, 2018

There are a couple methods that could be used here; I assume the intended method would be to use the chain rule (Method 2) below, but if you are willing to look at this from a different perspective, a more direct approach might be interesting.

Method 1: Reverse which of #x# and #y# are considered to be the independent and the dependent variables

The given equation: #y^2-x+1=0# can be written as a function with #x# as the dependent variable (the reverse of how we would normally look at this).
#color(white)("XXX")x_y=y^2+1#
This in effect exchanges the X and Y axis from what we are normally accustomed to.

The slope (in this reversed form) can be written directly (usind the exponent rule) as
#color(white)("XXX")(dx_y)/(dy) =2y#

#{: ("at "(2,-1),color(white)("xxxx"),"at " (2,1)), ("this becomes",,"this becomes"), ((dx_y)/(dy)=-2,,(dx_y)/(dy)=2), ("and using the slope-point form we have",,"and using the slope point form we have"), (x-2=-2(y-(-1)),,x-2=2(y-(-1))), ("or",,"or"), (x+2y=0,,x-2y=0) :}#

Method 2: Using the chain rule
#y^2-x+1=0#
#color(white)("XXX")rarr y=+-abs(x-1)^(1/2)#

If we let #u=x-1#
then
#{: ("when "y=+(x-1)^(1/2),color(white)("xxxx"),"when " y=-abs(x-1)^(1/2)), ("this case will apply for "(2,1),,"this case will apply for "(2,-1)), ((du)/(dx)=1" and "(dy)/(du)=1/2u^(-1/2),,(du)/(dx)=1" and "(dy)/(du)=-1/2u^(-1/2)), ((dy)/(dx)=(du)/(dx) * (dy)/(du),,(dy)/(dx)=(du)/(dx) * (dy)/(du)), (color(white)("XX")=1 * 1/(2sqrt(u)),,color(white)("XX")=1 * (-1/(2sqrt(u)))), (color(white)("XX")=1/(2sqrt(x-1)),,color(white)("XX")=-1/(2sqrt(x-1))), ("at "(2,1)" this becomes",,"at "(2,-1)" this becomes"), ((dy)/(dx)=1/2,,(dy)/(dx)=-1/2), ("using the slope-point form:",,"using the slope-point form:"), (y-1=(1/2) * (x-2),,y-1=(-1/2) * (x-2)), (-2y+2=x-2,,2y-2=x-2), (x+2y=0,,x-2y=0) :}#