At what point is the curve horizontal

the graph of x^2 y + xy^2 = 6

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

at one point, the curve is horizontal. find the exact coordinate of that point

any help would be greatly appreciated!!!

Since I concur with your derivative, I will begin there. The direction of the curve is horizontal where dy/dx is zero.

-(2xy + y²)/(x² + 2xy) = 0
2xy + y² = 0
y(2x + y) = 0
y = 0 or y = -2x

Substitute these into the curve equation.

x²y + xy² = 6

Let y = 0.
x²(0) + x(0)² = 6 ... No solution.

Let y = -2x.
x²(-2x) + x(-2x)² = 6
-2x³ + 4x³ = 6
2x³ = 6
x³ = 3
x = 3^(1/3)

Substitute it into the curve.
[3^(1/3)]²y + 3^(1/3)y² = 6
3^(2/3)y + 3^(1/3)y² = 6
3^(1/3)y² +3^(2/3)y - 6 = 0
y² +3^(1/3)y - 2·3^(2/3) = 0
[y - 3^(1/3)][y + 2·3^(1/3)] = 0
y = 3^(1/3) or y = -2·3^(1/3)

Reject the first solution for y because it does not satisfy the condition for dy/dx. We are left with one solution:

[3^(1/3), -2·3^(1/3)]

Looking at it again now, I cannot see why I did that last part with the substitution and factoring. I had x, and I had the relationship y = -2x. That was enough.

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The horizontal tangent occurs when you set the numerator of the derivative equal to zero and solve. So,
2xy+y^2=0
gives two solutions, either y=0 or y=2x. Since we cannot have the denominator zero, this is undefined, we exclude the origin. Hope that helps.