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.
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.
4 answers
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 through by y)
2x^2-x-1=0
what do I from here? It doesn't factor evenly. I'm stuck!
2x^2y-2xy+x-y=0
2x^2y-2xy+x=y
2x^2-2x+x=1 (divide through by y)
2x^2-x-1=0
what do I from here? It doesn't factor evenly. I'm stuck!
1 answer