Find the equation of the normal to the hyperbola `x^(2)-3y^(2)=144` at the positive end of the latus rectum.

Answer

To find the equation of the normal to the hyperbola \(x^2 - 3y^2 = 144\) at the positive end of the latus rectum, we can follow these steps:

### Step 1: Write the hyperbola in standard form
The given equation is:
\[ x^2 - 3y^2 = 144 \]
Dividing through by 144, we get:
\[ \frac{x^2}{144} - \frac{y^2}{48} = 1 \]
This is in the standard form of a hyperbola \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\), where \(a^2 = 144\) and \(b^2 = 48\).

### Step 2: Identify \(a\) and \(b\)
From the standard form, we find:
\[ a = 12 \quad \text{and} \quad b = \sqrt{48} = 4\sqrt{3} \]

### Step 3: Calculate the eccentricity \(e\)
The eccentricity \(e\) of a hyperbola is given by:
\[ e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{48}{144}} = \sqrt{1 + \frac{1}{3}} = \sqrt{\frac{4}{3}} = \frac{2}{\sqrt{3}} \]

### Step 4: Find the equation of the latus rectum
The latus rectum of the hyperbola is given by the equation \(x = ae\):
\[ x = 12 \cdot \frac{2}{\sqrt{3}} = \frac{24}{\sqrt{3}} = 8\sqrt{3} \]

### Step 5: Find the endpoints of the latus rectum
To find the endpoints of the latus rectum, substitute \(x = 8\sqrt{3}\) into the hyperbola equation:
\[ \frac{(8\sqrt{3})^2}{144} - \frac{y^2}{48} = 1 \]
Calculating:
\[ \frac{192}{144} - \frac{y^2}{48} = 1 \implies \frac{4}{3} - \frac{y^2}{48} = 1 \]
Rearranging gives:
\[ \frac{y^2}{48} = \frac{4}{3} - 1 = \frac{1}{3} \implies y^2 = 16 \implies y = \pm 4 \]
Thus, the endpoints of the latus rectum are \((8\sqrt{3}, 4)\) and \((8\sqrt{3}, -4)\).

### Step 6: Find the equation of the normal at the point \((8\sqrt{3}, 4)\)
The equation of the normal to the hyperbola at the point \((x_1, y_1)\) is given by:
\[ \frac{a^2 x}{x_1} + \frac{b^2 y}{y_1} = a^2 e^2 \]
Substituting \(a^2 = 144\), \(b^2 = 48\), \(e^2 = \frac{4}{3}\), and \((x_1, y_1) = (8\sqrt{3}, 4)\):
\[ \frac{144 x}{8\sqrt{3}} + \frac{48 y}{4} = 144 \cdot \frac{4}{3} \]
Simplifying:
\[ \frac{18\sqrt{3} x}{3} + 12y = 192 \]
\[ 6\sqrt{3} x + 12y = 192 \]
Dividing through by 6:
\[ \sqrt{3} x + 2y = 32 \]

### Final Equation
Thus, the equation of the normal to the hyperbola at the positive end of the latus rectum is:
\[ \sqrt{3} x + 2y = 32 \]