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

Find all points on the curve x^2y^2+xy=2 where the slope of tangent line is -1.

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 2:15pm
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.

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

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

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

## Answer This Question

## Related Questions

- Ap Calc AB - Determine (dy/dx) using implicit differentiation. cos(X^2Y^2) = x I...
- Calculus - Find y" by implicit differentiation. x^3+y^3 = 1 (x^3)'+(y^3)' = (1...
- Calculus - The line that is normal to the curve x^2=2xy-3y^2=0 at(1,1) ...
- Calculus - the curve: (x)(y^2)-(x^3)(y)=6 (dy/dx)=(3(x^2)y-(y^2))/(2xy-(x^3)) a...
- Calculus [finding slope of tangent line] - Find the slope of the tangent line to...
- calculus - Consider the curve defined by 2y^3+6X^2(y)- 12x^2 +6y=1 . a. Show ...
- calculus - Find the slope of the tangent line to the curve (2x+4y)^1/2+(2xy)^1/2...
- Calculus - Find all points on the curve x^2y^2+2xy=3 where the slope of the ...
- Calculus - Use implicit differentiation to find the slope of the tangent line to...
- Help Calc.! - original curve: 2y^3+6(x^2)y-12x^2+6y=1 dy/dx=(4x-2xy)/(x^2+y^2+1...

More Related Questions