Find the angle of rotation to eliminate xy term in the equation `x^(2)+2sqrt3xy-y^(2)=18.`

Answer

To find the angle of rotation that eliminates the \(xy\) term in the equation \(x^2 + 2\sqrt{3}xy - y^2 = 18\), we can follow these steps:

### Step 1: Rewrite the equation
We start by rewriting the given equation in a standard form:
\[ x^2 + 2\sqrt{3}xy - y^2 - 18 = 0 \]

### Step 2: Identify coefficients
Next, we identify the coefficients \(a\), \(b\), and \(h\) from the general conic section equation:
\[ Ax^2 + 2Hxy + By^2 + 2Gx + 2Fy + C = 0 \]
From our equation:
- \(A = 1\) (coefficient of \(x^2\))
- \(B = -1\) (coefficient of \(y^2\))
- \(2H = 2\sqrt{3}\) (coefficient of \(xy\)), hence \(H = \sqrt{3}\)

### Step 3: Calculate \(\theta\)
The formula to find the angle of rotation \(\theta\) to eliminate the \(xy\) term is given by:
\[ \tan(2\theta) = \frac{2H}{A - B} \]
Substituting the values we found:
- \(A = 1\)
- \(B = -1\)
- \(H = \sqrt{3}\)

We calculate:
\[ \tan(2\theta) = \frac{2\sqrt{3}}{1 - (-1)} = \frac{2\sqrt{3}}{1 + 1} = \frac{2\sqrt{3}}{2} = \sqrt{3} \]

### Step 4: Solve for \(\theta\)
Now, we find \(2\theta\):
\[ 2\theta = \tan^{-1}(\sqrt{3}) \]
The angle whose tangent is \(\sqrt{3}\) is \(60^\circ\) (or \(\frac{\pi}{3}\) radians). Therefore:
\[ \theta = \frac{60^\circ}{2} = 30^\circ \]

### Final Answer
Thus, the angle of rotation \(\theta\) to eliminate the \(xy\) term is:
\[ \theta = 30^\circ \quad \text{or} \quad \theta = \frac{\pi}{6} \text{ radians} \]

---