Question

Suppose x and y are real numbers and that $x^2 + 9y^2 - 4x + 6y + 4 = 0$ then the minimum value of (4x - 9y) is

Answer 1

6

Answer 2

The given equation can be rewritten by completing the square: $(x-2)^2 + 9(y+\frac{1}{3})^2 = 1$. Let $X = x-2$ and $Y = y+\frac{1}{3}$. The equation becomes $X^2 + 9Y^2 = 1$. The expression $4x - 9y$ can be rewritten in terms of X and Y as $4(X+2) - 9(Y-\frac{1}{3}) = 4X - 9Y + 11$. To find the minimum value, we parameterize the ellipse as $X = \cos\theta$ and $Y = \frac{1}{3}\sin\theta$. Substituting these into the expression gives $4\cos\theta - 9(\frac{1}{3}\sin\theta) + 11 = 4\cos\theta - 3\sin\theta + 11$. The minimum value of $4\cos\theta - 3\sin\theta$ is $-\sqrt{4^2 + (-3)^2} = -5$. Therefore, the minimum value of the expression is $-5 + 11 = 6$.