calculus
posted by Lee .
the curve with equation 2(x^2+y^2)^2=9(x^2y^2) is called a lemniscate. find the points on the lemniscate where the tangent is horizontal.

just plug and chug:
4(x^2 + y^2)(2x + 2yy') = 9(2x  2yy')
Now, the tangent is horizontal when y' = 0
4(x^2 + y^2)(2x) = 9(2x)
8x^3 + 8xy^2 = 18x
2x(4x^2 + 4y^2  9) = 0
So, either x=0
substitute back into original equation:
2y^4 = 9y^2
y=0
or,
4x^2 + 4y^2 = 9
x^2 = (9  4y^2)/4
Substitute that back into the original equation
2((9  4y^2)/4 + y^2)^2 = 9((9  4y^2)/4  y^2)
Expand the binomial and solve for y^2 
Tan^3(xy^2+y)=x
Use derivative 
Find .... Tan^3(xy^2+y)=x
Use derivative