The problem is to eliminate the cross-product term in the equation $x^2 + xy + y^2 = 6$ by rotating the axes. Then we put the equation in standard form. Finally, we graph the equation showing the rotated axes.

1. Problem Description

The problem is to eliminate the cross-product term in the equation $x^2 + xy + y^2 = 6$ by rotating the axes. Then we put the equation in standard form. Finally, we graph the equation showing the rotated axes.

2. Solution Steps

First, we identify the coefficients in the general equation $Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$. In this case, $A = 1$, $B = 1$, $C = 1$, $D = 0$, $E = 0$, and $F = -6$.

To eliminate the $xy$ term, we need to rotate the axes by an angle $\theta$ such that

$$\cot(2\theta) = \frac{A - C}{B}$$

In our case, $A = 1$, $B = 1$, and $C = 1$, so

$$\cot(2\theta) = \frac{1 - 1}{1} = 0$$

Therefore, $2\theta = \frac{\pi}{2}$, which means $\theta = \frac{\pi}{4}$.

The rotation formulas are:

$$x = x' \cos\theta - y' \sin\theta$$

$$y = x' \sin\theta + y' \cos\theta$$

Since $\theta = \frac{\pi}{4}$, we have $\cos\theta = \sin\theta = \frac{\sqrt{2}}{2}$. Thus,

$$x = \frac{\sqrt{2}}{2} (x' - y')$$

$$y = \frac{\sqrt{2}}{2} (x' + y')$$

Substituting these expressions into the given equation $x^2 + xy + y^2 = 6$, we get

$$(\frac{\sqrt{2}}{2} (x' - y'))^2 + (\frac{\sqrt{2}}{2} (x' - y'))(\frac{\sqrt{2}}{2} (x' + y')) + (\frac{\sqrt{2}}{2} (x' + y'))^2 = 6$$

$$\frac{1}{2} (x'^2 - 2x'y' + y'^2) + \frac{1}{2} (x'^2 - y'^2) + \frac{1}{2} (x'^2 + 2x'y' + y'^2) = 6$$

$$\frac{1}{2} x'^2 - x'y' + \frac{1}{2} y'^2 + \frac{1}{2} x'^2 - \frac{1}{2} y'^2 + \frac{1}{2} x'^2 + x'y' + \frac{1}{2} y'^2 = 6$$

Combining like terms, we have

$$\frac{3}{2} x'^2 + \frac{1}{2} y'^2 = 6$$

Multiplying by 2, we get

$$3x'^2 + y'^2 = 12$$

Dividing by 12, we obtain the standard form:

$$\frac{x'^2}{4} + \frac{y'^2}{12} = 1$$

This is an ellipse with semi-major axis $a = \sqrt{12} = 2\sqrt{3}$ along the $y'$-axis and semi-minor axis $b = \sqrt{4} = 2$ along the $x'$-axis.

3. Final Answer

The equation in standard form after rotation is $\frac{x'^2}{4} + \frac{y'^2}{12} = 1$.

This is an ellipse centered at the origin, rotated by $\theta = \frac{\pi}{4}$ from the original $x, y$ axes. The semi-major axis is $2\sqrt{3}$ along the $y'$ axis and the semi-minor axis is $2$ along the $x'$ axis.