Posted by Riley on Thursday, April 2, 2009 at 1:14pm.
You can insert dy/dx = -1 at this point:
x^2*2ydy/dx + 2x*y^2 + x*dy/dx + y =0
You then get an equation relating x and y. You also know that x and y are on te curve, so they satisfy the equation of the curve:
x^2y^2+xy = 2
So, you have two equations for the two unknowns x and y.
So I put -1 in for dy/dx and got this equation: -2x^2y+2xy-x+y=0. Then would I solve for either x or y and substitute back in to the original equation to get a value for x and y?
Yes.
I got this for y.
2x^2y-2xy+x-y=0
2x^2y-2xy+x=y
2x^2-2x+x=1 (divide thru by y)
2x^2-x-1=0
what do I from here? It doesn't factor evenly. I'm stuck!
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