To put the given conic section into standard form, we need to remove the cross term $(-2xy)$ by rotating the coordinate axes and translating the origin.

The given equation is:

$2x^2 - 2xy + 2y^2 + 4x + 6y + 3 = 0.$

Step 1: Identify the rotation angle

A general conic section is represented by:

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

In our equation:

• $A = 2$
• $B = -2$
• $C = 2$
• $D = 4$
• $E = 6$
• $F = 3$

To eliminate the $xy$ term, we rotate the coordinate system by an angle $\theta$ such that:

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

Substituting the values:

$\tan(2\theta) = \frac{-2}{2 - 2} = \frac{-2}{0} \implies \tan(2\theta) = \infty.$

Since $\tan(2\theta) = \infty$, $2\theta = 90^\circ$, which gives $\theta = 45^\circ$. Therefore, we should rotate the axes by $45^\circ$.

Step 2: Apply the rotation transformation

The rotation transformation equations are:

$x = x' \cos(\theta) - y' \sin(\theta), \quad y = x' \sin(\theta) + y' \cos(\theta).$

For $\theta = 45^\circ$:

$x = \frac{x' - y'}{\sqrt{2}}, \quad y = \frac{x' + y'}{\sqrt{2}}.$

Now, substitute these into the original equation:

$2 \left( \frac{x' - y'}{\sqrt{2}} \right)^2 - 2 \left( \frac{x' - y'}{\sqrt{2}} \right) \left( \frac{x' + y'}{\sqrt{2}} \right) + 2 \left( \frac{x' + y'}{\sqrt{2}} \right)^2 + 4 \left( \frac{x' - y'}{\sqrt{2}} \right) + 6 \left( \frac{x' + y'}{\sqrt{2}} \right) + 3 = 0.$

Simplifying each term:

$2 \cdot \frac{(x' - y')^2}{2} - 2 \cdot \frac{(x'^2 - y'^2)}{2} + 2 \cdot \frac{(x' + y')^2}{2} + 4 \cdot \frac{x' - y'}{\sqrt{2}} + 6 \cdot \frac{x' + y'}{\sqrt{2}} + 3 = 0.$

Combine like terms:

$(x' - y')^2 - (x'^2 - y'^2) + (x' + y')^2 + 2\sqrt{2}(x' + y') + 3 = 0.$

Expanding the squares:

$(x'^2 - 2x'y' + y'^2) - (x'^2 - y'^2) + (x'^2 + 2x'y' + y'^2) + 2\sqrt{2}(x' + y') + 3 = 0.$

Combine like terms:

$x'^2 + y'^2 + 2\sqrt{2}x' + 2\sqrt{2}y' + 3 = 0.$

Step 3: Translate the axes

To further simplify, we perform a translation by setting:

$x'' = x' + \frac{\sqrt{2}}{2}, \quad y'' = y' + \frac{\sqrt{2}}{2}.$

Substitute into the equation:

$(x'' - \frac{\sqrt{2}}{2})^2 + (y'' - \frac{\sqrt{2}}{2})^2 + 2\sqrt{2} \left( x'' - \frac{\sqrt{2}}{2} \right) + 2\sqrt{2} \left( y'' - \frac{\sqrt{2}}{2} \right) + 3 = 0.$

Expanding and simplifying:

$x''^2 + y''^2 - 1 = 0.$

Step 4: Identify the type of conic section

The simplified equation is:

$x''^2 + y''^2 = 1.$

This is the equation of a circle with radius 1 centered at the origin.

Conclusion

The given conic section is a circle.

Would you like more details on any of these steps, or do you have any questions?

Additional Related Questions

1. What are the general forms of equations for different types of conic sections?
2. How does one determine the angle of rotation for eliminating the cross-term in a conic section?
3. What is the geometric interpretation of translating the axes in conic sections?
4. How can the discriminant of a conic section equation determine its type?
5. What are some real-world applications of conic sections?

Tip: Rotating and translating axes is a powerful method to identify the type of conic section, and it simplifies their analysis in geometry.