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Consider the conic section

2x2-2xy+2y2+4x+6y+3=0.

Rotate and translate the axes to write it in standard form. Hence name the type

of conic section.

Let's solve the conic section given by the equation:

\[
2x^2 - 2xy + 2y^2 + 4x + 6y + 3 = 0
\]

### Step 1: Rotation of Axes

The presence of the \(xy\) term indicates that the conic section is rotated with respect to the standard axes. To eliminate the \(xy\) term, we need to apply a rotation of axes. The angle \(\theta\) required for this transformation is found by:

\[
\tan(2\theta) = \frac{B}{A - C}
\]

In the general conic form \(Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0\), the coefficients are:
- \(A = 2\)
- \(B = -2\)
- \(C = 2\)

Substitute into the equation for \(\tan(2\theta)\):

\[
\tan(2\theta) = \frac{-2}{2 - 2} = \frac{-2}{0}
\]

This implies that \(2\theta = -\frac{\pi}{2}\), so \(\theta = -\frac{\pi}{4}\).

### Step 2: Perform the Rotation

Next, apply the rotation formulas:

\[
x = x'\cos\theta - y'\sin\theta
\]
\[
y = x'\sin\theta + y'\cos\theta
\]

For \(\theta = -\frac{\pi}{4}\), we have \(\cos\theta = \frac{\sqrt{2}}{2}\) and \(\sin\theta = -\frac{\sqrt{2}}{2}\). Substituting into the equations:

\[
x = \frac{\sqrt{2}}{2}(x' + y')
\]
\[
y = \frac{\sqrt{2}}{2}(-x' + y')
\]

Substituting these into the original conic equation:

\[
2x^2 - 2xy + 2y^2 + 4x + 6y + 3 = 0
\]

After rotation, this will yield a new equation in terms of \(x'\) and \(y'\), where the cross term \(x'y'\) should be eliminated.

### Step 3: Translation of Axes

Once the conic is rotated, the next step is to translate the axes to complete the square and write the equation in standard form. After that, we can determine the type of conic section by comparing the coefficients.

Let me calculate the rotated and translated equation next.After rotating the axes by \(-\frac{\pi}{4}\), the new equation becomes:

\[
3x'^2 - \sqrt{2}x' + y'^2 + 5\sqrt{2}y' + 3 = 0
\]

### Step 4: Translation of Axes

To simplify this further, we complete the square for both \(x'\) and \(y'\) terms.

#### For \(x'\):

We have the term \(3x'^2 - \sqrt{2}x'\). Completing the square:

\[
3 \left(x' - \frac{\sqrt{2}}{6}\right)^2 - \frac{1}{6}
\]

#### For \(y'\):

We have the term \(y'^2 + 5\sqrt{2}y'\). Completing the square:

\[
\left(y' + \frac{5\sqrt{2}}{2}\right)^2 - \frac{25}{2}
\]

Substitute these into the equation:

\[
3 \left(x' - \frac{\sqrt{2}}{6}\right)^2 + \left(y' + \frac{5\sqrt{2}}{2}\right)^2 - \frac{1}{6} - \frac{25}{2} + 3 = 0
\]

Simplifying:

\[
3 \left(x' - \frac{\sqrt{2}}{6}\right)^2 + \left(y' + \frac{5\sqrt{2}}{2}\right)^2 = \frac{61}{6}
\]

### Step 5: Identify the Conic Section

This is now in the standard form of an **ellipse**, since both \(x'^2\) and \(y'^2\) terms have positive coefficients. Therefore, the given equation represents an **ellipse**.

Would you like further details on how the translation was done, or help with a similar topic?

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Here are 5 related questions you might be interested in:
1. How is the rotation of axes derived for conic sections?
2. What are the general forms of different types of conic sections?
3. How do we classify conic sections based on their general equation?
4. What are the applications of conic sections in real-life problems?
5. How can we translate conic sections to new coordinates without rotation?

Tip: Always check for the presence of an \(xy\) term in a conic section equation, as it signals a rotated conic.

Consider the conic section 2x^2 - 2xy + 2y^2 + 4x + 6y + 3 = 0. Rotate and translate the axes to write it in standard form. Hence name the type of conic section.

Learn how to solve the conic section equation 2x^2 - 2xy + 2y^2 + 4x + 6y + 3 = 0 by rotating and translating the axes. This problem involves applying the rotation of axes formula and completing the square to write the equation in standard form. The final form reveals the conic section as an ellipse. Suitable for advanced high school or undergraduate students.

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