Transform to axes inclined at 45° to the original axes the equations
`x^(2) - y^(2) = a^(2)`

Answer

To transform the equation \( x^2 - y^2 = a^2 \) to axes inclined at 45° to the original axes, we follow these steps:

### Step 1: Define the Transformation
We need to express the original coordinates \( (x, y) \) in terms of the new coordinates \( (x', y') \). The transformation for axes inclined at an angle \( \theta \) is given by:
\[ x = x' \cos \theta - y' \sin \theta \]
\[ y = x' \sin \theta + y' \cos \theta \]
For \( \theta = 45^\circ \), we have \( \cos 45^\circ = \sin 45^\circ = \frac{1}{\sqrt{2}} \).

### Step 2: Substitute the Values
Substituting \( \theta = 45^\circ \) into the transformation equations, we get:
\[ x = \frac{x'}{\sqrt{2}} - \frac{y'}{\sqrt{2}} \]
\[ y = \frac{x'}{\sqrt{2}} + \frac{y'}{\sqrt{2}} \]

### Step 3: Substitute into the Original Equation
Now, we substitute these expressions for \( x \) and \( y \) into the original equation \( x^2 - y^2 = a^2 \):
\[ \left( \frac{x'}{\sqrt{2}} - \frac{y'}{\sqrt{2}} \right)^2 - \left( \frac{x'}{\sqrt{2}} + \frac{y'}{\sqrt{2}} \right)^2 = a^2 \]

### Step 4: Expand the Squares
Expanding both squares:
\[ \left( \frac{x'}{\sqrt{2}} - \frac{y'}{\sqrt{2}} \right)^2 = \frac{x'^2}{2} - \frac{x'y'}{2} + \frac{y'^2}{2} \]
\[ \left( \frac{x'}{\sqrt{2}} + \frac{y'}{\sqrt{2}} \right)^2 = \frac{x'^2}{2} + \frac{x'y'}{2} + \frac{y'^2}{2} \]

### Step 5: Combine the Results
Now substituting back into the equation:
\[ \left( \frac{x'^2}{2} - \frac{x'y'}{2} + \frac{y'^2}{2} \right) - \left( \frac{x'^2}{2} + \frac{x'y'}{2} + \frac{y'^2}{2} \right) = a^2 \]
This simplifies to:
\[ -\frac{x'y'}{2} - \frac{x'y'}{2} = a^2 \]
\[ - x'y' = 2a^2 \]

### Step 6: Rearranging the Equation
Rearranging gives us:
\[ 2x'y' + a^2 = 0 \]

### Final Result
Thus, the transformed equation is:
\[ 2x'y' + a^2 = 0 \]

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