Could someone else check my work? I can't find my mistake.
Function : x^2 y + x y^2 = 2x
two points of two tangent lines (1,1) and (1,-2)

Question

Could someone else check my work? I can't find my mistake.
Function : x^2 y + x y^2 = 2x 
two points of two tangent lines (1,1) and (1,-2)

anonymous · Answer

First I find the derivative of x^2 y + x y^2 = 2x
which is $$y' = \frac{ 2-2xy-y^2 }{ x^2+2xy }$$

anonymous · Answer

Next I find the slopes of both tangent lines to form an equation for both lines.

anonymous · Answer

I plug in (1,1) into y' and get the slope of -1/3.
I plug in (1,-2) into y' and get the slope of -2/3

anonymous · Answer

Now I form an equation with the points and slopes I found.

anonymous · Answer

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

anonymous · Answer

$$y = \frac{ -x + 4 }{ 3 }                     , y= \frac{- 2x-4 }{3 }$$

anonymous · Answer

Then I proceed to find x and y where they intersect each other.

anonymous · Answer

(-x+4)/3 = (-2x-4)/3
 3(-x+4) = 3(-2x-4)
  -3x + 12 = -6x - 12
    3 x = -24
     x = -8

anonymous · Answer

y = (-(-8)+4)/3
y = 12/3
 y = 4