Transform to axes inclined at 30° to the original axes the equation `x^(2) + 2 sqrt"" 3 xy - y^(2) = 2a^(2)`

Answer

To transform the given equation \( x^2 + 2\sqrt{3}xy - y^2 = 2a^2 \) to axes inclined at \( 30^\circ \) to the original axes, we will follow these steps:

### Step 1: Identify the transformation equations
For axes inclined at an angle \( \theta \) (in this case, \( 30^\circ \)), the transformations for \( x \) and \( y \) in terms of the new coordinates \( x' \) and \( y' \) are given by:
\[ x = x' \cos \theta - y' \sin \theta \]
\[ y = x' \sin \theta + y' \cos \theta \]
Substituting \( \theta = 30^\circ \):
\[ \cos 30^\circ = \frac{\sqrt{3}}{2}, \quad \sin 30^\circ = \frac{1}{2} \]
Thus, the transformations become:
\[ x = \frac{\sqrt{3}}{2} x' - \frac{1}{2} y' \]
\[ y = \frac{1}{2} x' + \frac{\sqrt{3}}{2} y' \]

### Step 2: Substitute the transformations into the original equation
Now we substitute these expressions for \( x \) and \( y \) into the original equation:
\[ x^2 + 2\sqrt{3}xy - y^2 = 2a^2 \]

Substituting for \( x \):
\[ x^2 = \left(\frac{\sqrt{3}}{2} x' - \frac{1}{2} y'\right)^2 = \frac{3}{4} x'^2 - \frac{\sqrt{3}}{2} x'y' + \frac{1}{4} y'^2 \]

Substituting for \( y \):
\[ y^2 = \left(\frac{1}{2} x' + \frac{\sqrt{3}}{2} y'\right)^2 = \frac{1}{4} x'^2 + \frac{\sqrt{3}}{2} x'y' + \frac{3}{4} y'^2 \]

Substituting for \( xy \):
\[ xy = \left(\frac{\sqrt{3}}{2} x' - \frac{1}{2} y'\right)\left(\frac{1}{2} x' + \frac{\sqrt{3}}{2} y'\right) = \frac{\sqrt{3}}{4} x'^2 + \frac{3}{4} y'^2 - \frac{1}{4} x'y' \]

### Step 3: Substitute these into the equation
Now substituting these into the equation:
\[ \left(\frac{3}{4} x'^2 - \frac{\sqrt{3}}{2} x'y' + \frac{1}{4} y'^2\right) + 2\sqrt{3}\left(\frac{\sqrt{3}}{4} x'^2 + \frac{3}{4} y'^2 - \frac{1}{4} x'y'\right) - \left(\frac{1}{4} x'^2 + \frac{\sqrt{3}}{2} x'y' + \frac{3}{4} y'^2\right) = 2a^2 \]

### Step 4: Simplify the equation
Combine like terms:
1. For \( x'^2 \):
\[ \frac{3}{4} x'^2 + \frac{3}{2} x'^2 - \frac{1}{4} x'^2 = \frac{3}{4} x'^2 + \frac{6}{4} x'^2 - \frac{1}{4} x'^2 = \frac{8}{4} x'^2 = 2 x'^2 \]

2. For \( y'^2 \):
\[ \frac{1}{4} y'^2 + \frac{3\sqrt{3}}{2} y'^2 - \frac{3}{4} y'^2 = \frac{1}{4} y'^2 + \frac{6}{4} y'^2 - \frac{3}{4} y'^2 = \frac{4}{4} y'^2 = y'^2 \]

3. For \( x'y' \):
\[ -\frac{\sqrt{3}}{2} x'y' - \frac{1}{2} x'y' - \frac{\sqrt{3}}{2} x'y' = -\left(\sqrt{3} + \frac{1}{2}\right)x'y' = -\left(\frac{2\sqrt{3} + 1}{2}\right)x'y' \]

### Step 5: Final equation
Putting it all together, we get:
\[ 2x'^2 - \left(\frac{2\sqrt{3} + 1}{2}\right)x'y' - y'^2 = 2a^2 \]

### Step 6: Rearranging
Rearranging gives us the transformed equation:
\[ 2x'^2 - y'^2 - \left(\frac{2\sqrt{3} + 1}{2}\right)x'y' = 2a^2 \]

### Final Result
The transformed equation in the new coordinates is:
\[ 2x'^2 - y'^2 - \left(\frac{2\sqrt{3} + 1}{2}\right)x'y' = 2a^2 \]

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