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October 31, 2014

October 31, 2014

Posted by **Riley** on Thursday, April 2, 2009 at 1:14pm.

This is what I got.

x^2y^2+xy=2

x^2*2ydy/dx + 2x*y^2 + x*dy/dx + y =0

dy/dx(2yx^2+x)= -2xy^2-y

dy/dx = (-2xy^2-y)/(x+2yx^2)

how do I finish from here? I need to know where the slop of the tangent line equals -1.

- calculus -
**Count Iblis**, Thursday, April 2, 2009 at 2:15pmYou 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.

- calculus -
**Riley**, Thursday, April 2, 2009 at 2:26pmSo 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?

- calculus -
**Count Iblis**, Thursday, April 2, 2009 at 5:14pmYes.

- calculus -
**Anonymous**, Thursday, April 2, 2009 at 8:26pmI 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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