Find the equation of the normal at `theta=(pi)/(3)` to the hyperbola `3x^(2)-4y^(2)=12`.

Answer

To find the equation of the normal at \(\theta = \frac{\pi}{3}\) to the hyperbola \(3x^2 - 4y^2 = 12\), we will follow these steps:

### Step 1: Rewrite the hyperbola in standard form
We start with the equation of the hyperbola:
\[ 3x^2 - 4y^2 = 12 \]
We divide the entire equation by 12 to put it in standard form:
\[ \frac{x^2}{4} - \frac{y^2}{3} = 1 \]
Here, we identify \(A^2 = 4\) and \(B^2 = 3\).

### Step 2: Identify parameters \(A\) and \(B\)
From the standard form, we can find:
\[ A = \sqrt{4} = 2 \quad \text{and} \quad B = \sqrt{3} \]

### Step 3: Use the equation of the normal
The equation of the normal to the hyperbola is given by:
\[ \frac{Ax}{\sec \theta} + \frac{By}{\tan \theta} = A^2 + B^2 \]
Substituting \(A\) and \(B\) into the equation, we have:
\[ \frac{2x}{\sec \theta} + \frac{\sqrt{3}y}{\tan \theta} = 4 + 3 = 7 \]

### Step 4: Substitute \(\theta = \frac{\pi}{3}\)
Now, we substitute \(\theta = \frac{\pi}{3}\):
- \(\sec \frac{\pi}{3} = 2\)
- \(\tan \frac{\pi}{3} = \sqrt{3}\)

Substituting these values into the normal equation:
\[ \frac{2x}{2} + \frac{\sqrt{3}y}{\sqrt{3}} = 7 \]
This simplifies to:
\[ x + y = 7 \]

### Step 5: Final equation of the normal
Thus, the equation of the normal at \(\theta = \frac{\pi}{3}\) to the hyperbola is:
\[ x + y = 7 \]