The problem asks us to eliminate the $xy$ term from the given equation $x^2 + xy + y^2 = 6$ by rotating the axes. We then need to express the equation in standard form and finally graph the equation showing the rotated axes.

1. Problem Description

The problem asks us to eliminate the $xy$ term from the given equation $x^2 + xy + y^2 = 6$ by rotating the axes. We then need to express the equation in standard form and finally graph the equation showing the rotated axes.

2. Solution Steps

The general form of a conic section is given by

$Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$.

In our 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

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

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

$2\theta = \frac{\pi}{2}$

$\theta = \frac{\pi}{4}$

The rotation equations are:

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

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

Since $\theta = \frac{\pi}{4}$, we have $\cos\theta = \sin\theta = \frac{1}{\sqrt{2}}$.

So, $x = \frac{1}{\sqrt{2}}(x' - y')$ and $y = \frac{1}{\sqrt{2}}(x' + y')$.

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

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

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

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

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

Dividing by 6, we get:

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

This is an ellipse with $a^2 = 12$ and $b^2 = 4$, so $a = \sqrt{12} = 2\sqrt{3}$ and $b = 2$.

3. Final Answer

The equation in standard form is $\frac{x'^2}{4} + \frac{y'^2}{12} = 1$. This is an ellipse centered at the origin, with major axis along the $y'$-axis and minor axis along the $x'$-axis. The angle of rotation is $\theta = \frac{\pi}{4}$.