For the curve given by 4x^2+y^2=48+2xy, find the positive y-coordinate given that the x-coordinate is 2.

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To find the positive y-coordinate on the curve when the x-coordinate is 2, we can start by substituting the given value of x into the equation and solving for y.

The equation of the curve is 4x^2 + y^2 = 48 + 2xy.

Substituting x = 2 into the equation, we get:

4(2)^2 + y^2 = 48 + 2(2)(y)
16 + y^2 = 48 + 4y

Rearranging terms, we have:

y^2 - 4y + 32 = 0

Now, we can solve this quadratic equation using factoring, completing the square, or the quadratic formula. In this case, let's use the quadratic formula:

y = (-b ± √(b^2 - 4ac)) / (2a)

For our equation, a = 1, b = -4, and c = 32. Substituting those values:

y = (-(-4) ± √((-4)^2 - 4(1)(32))) / (2 * 1)
y = (4 ± √(16 - 128)) / 2
y = (4 ± √(-112)) / 2

Since we are looking for a positive y-coordinate on the curve, we can ignore the negative square root because it would yield a negative y-value.

Therefore, the positive y-coordinate when x = 2 is:

y = (4 + √(-112)) / 2

The value inside the square root, -112, is negative, so this curve does not have any real solutions for the positive y-coordinate when x = 2.