Find the equation of the normal to the hyperbola `4x^(2) - 9y^(2) = 144` at the point whose eccentric angle `theta` is `pi//3`

Answer

To find the equation of the normal to the hyperbola \(4x^2 - 9y^2 = 144\) at the point whose eccentric angle \(\theta\) is \(\frac{\pi}{3}\), we can follow these steps:

### Step 1: Rewrite the hyperbola in standard form
The given hyperbola is \(4x^2 - 9y^2 = 144\). We can divide the entire equation by 144 to rewrite it in standard form:

\[ \frac{x^2}{36} - \frac{y^2}{16} = 1 \]

This shows that \(a^2 = 36\) and \(b^2 = 16\). Therefore, \(a = 6\) and \(b = 4\).

### Step 2: Find the coordinates of the point on the hyperbola
Using the eccentric angle \(\theta = \frac{\pi}{3}\), we can find the coordinates of the point \(P\) on the hyperbola using the formulas:

\[ x = a \sec \theta, \quad y = b \tan \theta \]

Calculating these values:

\[ x = 6 \sec\left(\frac{\pi}{3}\right) = 6 \cdot 2 = 12 \]
\[ y = 4 \tan\left(\frac{\pi}{3}\right) = 4 \cdot \sqrt{3} = 4\sqrt{3} \]

Thus, the point \(P\) is \((12, 4\sqrt{3})\).

### Step 3: Use the normal equation formula
The equation of the normal to the hyperbola at the point \((x_1, y_1)\) is given by:

\[ \frac{a^2 x}{x_1} + \frac{b^2 y}{y_1} = a^2 + b^2 \]

Substituting \(a^2 = 36\), \(b^2 = 16\), \(x_1 = 12\), and \(y_1 = 4\sqrt{3}\):

\[ \frac{36x}{12} + \frac{16y}{4\sqrt{3}} = 36 + 16 \]

### Step 4: Simplify the equation
Calculating the left-hand side:

\[ \frac{36x}{12} = 3x \]
\[ \frac{16y}{4\sqrt{3}} = \frac{4y}{\sqrt{3}} \]

The right-hand side:

\[ 36 + 16 = 52 \]

Thus, the equation becomes:

\[ 3x + \frac{4y}{\sqrt{3}} = 52 \]

### Step 5: Rearranging the equation
To express this in a more standard form, we can multiply through by \(\sqrt{3}\) to eliminate the fraction:

\[ 3\sqrt{3}x + 4y = 52\sqrt{3} \]

### Final Answer
The equation of the normal to the hyperbola at the given point is:

\[ 3\sqrt{3}x + 4y = 52\sqrt{3} \]