Find the angle through which the axes be rotated to remove the xy term from the equation
`ax^(2)+2hxy+ay^(2) = 0`

Answer

To find the angle through which the axes should be rotated to remove the xy term from the equation \( ax^2 + 2hxy + ay^2 = 0 \), we can follow these steps:

### Step 1: Understand the Rotation of Axes
The rotation of axes involves transforming the coordinates \( (x, y) \) into new coordinates \( (x', y') \) using the following formulas:
\[ x = x' \cos \theta - y' \sin \theta \]
\[ y = x' \sin \theta + y' \cos \theta \]
where \( \theta \) is the angle of rotation.

### Step 2: Substitute the New Coordinates
Substituting these expressions into the original equation \( ax^2 + 2hxy + ay^2 = 0 \):
\[ a(x' \cos \theta - y' \sin \theta)^2 + 2h(x' \cos \theta - y' \sin \theta)(x' \sin \theta + y' \cos \theta) + a(x' \sin \theta + y' \cos \theta)^2 = 0 \]

### Step 3: Expand the Equation
Now, we will expand each term:
1. \( a(x' \cos \theta - y' \sin \theta)^2 = a(x'^2 \cos^2 \theta - 2x'y' \cos \theta \sin \theta + y'^2 \sin^2 \theta) \)
2. \( a(x' \sin \theta + y' \cos \theta)^2 = a(x'^2 \sin^2 \theta + 2x'y' \sin \theta \cos \theta + y'^2 \cos^2 \theta) \)
3. The mixed term \( 2h(x' \cos \theta - y' \sin \theta)(x' \sin \theta + y' \cos \theta) \) expands to:
\[ 2h(x'y' \cos^2 \theta - x'y' \sin^2 \theta - y'x' \sin \theta \cos \theta + y'x' \sin \theta \cos \theta) = 2h(x'y'(\cos^2 \theta - \sin^2 \theta)) \]

### Step 4: Collect Like Terms
After expanding, we will collect the coefficients of \( x'y' \):
\[ x'y' \left(-2a \cos \theta \sin \theta + 2h(\cos^2 \theta - \sin^2 \theta)\right) = 0 \]

### Step 5: Set the Coefficient to Zero
To remove the \( x'y' \) term, we set the coefficient to zero:
\[ -2a \cos \theta \sin \theta + 2h(\cos^2 \theta - \sin^2 \theta) = 0 \]
Dividing by 2:
\[ -a \cos \theta \sin \theta + h(\cos^2 \theta - \sin^2 \theta) = 0 \]

### Step 6: Use Trigonometric Identities
Using the identity \( \cos^2 \theta - \sin^2 \theta = \cos 2\theta \) and \( \sin \theta \cos \theta = \frac{1}{2} \sin 2\theta \):
\[ -a \cdot \frac{1}{2} \sin 2\theta + h \cos 2\theta = 0 \]
Rearranging gives:
\[ h \cos 2\theta = \frac{a}{2} \sin 2\theta \]

### Step 7: Form the Tangent Equation
This can be rearranged to form:
\[ \tan 2\theta = \frac{2h}{a} \]

### Step 8: Solve for \( \theta \)
To find \( \theta \), we take the arctangent:
\[ 2\theta = \tan^{-1}\left(\frac{2h}{a}\) \]
Thus,
\[ \theta = \frac{1}{2} \tan^{-1}\left(\frac{2h}{a}\right) \]

### Final Answer
The angle through which the axes should be rotated to remove the \( xy \) term is:
\[ \theta = \frac{1}{2} \tan^{-1}\left(\frac{2h}{a}\right) \]

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