Originally Posted by

**topsquark** $\displaystyle xy {\color{red}+} 2x - y = 0$ Mr F says: Dang it!!

$\displaystyle (x - 1)y = 2x$

$\displaystyle y = \frac{2x}{x - 1}$

Now find the slope of the line tangent to this curve at (x, y):

$\displaystyle y' = \frac{-2}{(x - 1)^2}$

The normal to the tangent line has the slope

$\displaystyle -\frac{1}{y'} = -\frac{(x - 1)^2}{-2} = \frac{1}{2}x^2 - x + \frac{1}{2}$

So given a point (x, y) on the original curve you now know the slope of the line normal to the curve at (x, y). I'll let you finish it out from here.

-Dan