Problem 34 - Algebra and Trigonometry

Question

Identify each equation without applying a rotation of axes. $$3 x^{2}-2 \sqrt{3} x y+y^{2}+2 x+2 \sqrt{3} y=0$$

Step-by-Step Solution

Answer

The equation is an ellipse with its center shifted, not rotated. The standard form is \(3(x + \frac{1}{3})^2 - 2\sqrt{3}(x + \frac{1}{3})(y + \sqrt{3}) + (y + \sqrt{3})^2 = \frac{10}{9}\).

Step 1: Regroup the equation.

Rewrite the given equation by grouping the \(x\) and \(y\)-terms together. This results in: \(3x^2 + 2x - 2\sqrt{3}xy + y^2 + 2\sqrt{3}y = 0\)

Step 2: Complete the square for the x-terms.

To complete the square, halve the coefficient of the \(x\)-term, square it and add it to both sides of the equation. For this equation, the \(x\)-terms can be rewritten as: \(3(x^2 + \frac{2}{3}x + \frac{1}{9}) - 2\sqrt{3}xy + y^2 + 2\sqrt{3}y = \frac{1}{9}\)

Step 3: Complete the square for the y-terms.

Using the same process as in Step 2, complete the square for the \(y\)-terms. The equation then becomes: \(3(x + \frac{1}{3})^2 - 2\sqrt{3}xy + (y^2 + 2\sqrt{3}y + 3) = \frac{1}{9} + 3\)

Step 4: Rewrite the equation in standard form.

Once the squares are completed, rewrite the equation to resemble the standard form of an ellipse equation. The final result will be: \(3(x + \frac{1}{3})^2 - 2\sqrt{3}(x + \frac{1}{3})(y + \sqrt{3}) + (y + \sqrt{3})^2 = \frac{10}{9}\)