Find the slope of the tangent line to the curve 2(x^2+y^2)^2=25(x^2−y^2) at the point (−3,−1)?

Question

Here's what I did:

2(x^4 + y^4) = 25(x^2-y^2)
2x^4 + 2y^4 = 25x^2 - 25y^2
8x^3 + 8y^3(dy/dx) = 50x - 50y(dy/dx)
d/dx(8y^3 + 50y) = 50x - 8x^3
d/dx = (50x-8x^3)/(8y^3 + 50y)

and got m= -1.13793

What did I do wrong?

Answer 1

in algebra I you learned that (a+b)^2 ≠ a^2 + b^2

That's still true, even if you're taking pre-cal! :-)

2(x^2+y^2)^2=25(x^2−y^2)
2 * 2(x^2+y^2) (2x+2yy') = 25(2x-2yy')
y' =
-x(4x^2+4y^2-25)
--------------------------
y(4x^2+4y^2+25)

So, at (-3,1), y' = -9/13

Also, don't lose the y in your dy/dx

Answer 2

thanks...though I asked for (-3,-1).