Equation of the tangent to the hyperbola `4x^(2)-9y^(2)=1` with eccentric angle `pi//6` is

Options

A. `4x+3y=sqrt3`
B. `4x-3y=sqrt3`
C. `3x-4y=sqrt3`
D. `3x-4y=sqrt5`

Answer

To find the equation of the tangent to the hyperbola given by \(4x^2 - 9y^2 = 1\) at the eccentric angle \(\frac{\pi}{6}\), we will follow these steps:

### Step 1: Convert the hyperbola to standard form
The given equation of the hyperbola is:
\[ 4x^2 - 9y^2 = 1 \]
We can rewrite this in the standard form of a hyperbola:
\[ \frac{x^2}{\left(\frac{1}{2}\right)^2} - \frac{y^2}{\left(\frac{1}{3}\right)^2} = 1 \]
Thus, we have:
\[ a = \frac{1}{2}, \quad b = \frac{1}{3} \]

### Step 2: Use the formula for the equation of the tangent
The equation of the tangent to the hyperbola at an eccentric angle \(\theta\) is given by:
\[ \frac{x}{a} \sec \theta - \frac{y}{b} \tan \theta = 1 \]
Substituting \(a\) and \(b\) into the equation, we have:
\[ \frac{x}{\frac{1}{2}} \sec \theta - \frac{y}{\frac{1}{3}} \tan \theta = 1 \]

### Step 3: Substitute the eccentric angle
Given that \(\theta = \frac{\pi}{6}\), we find:
\[ \sec \frac{\pi}{6} = \frac{2}{\sqrt{3}}, \quad \tan \frac{\pi}{6} = \frac{1}{\sqrt{3}} \]
Now substituting these values into the tangent equation:
\[ \frac{x}{\frac{1}{2}} \cdot \frac{2}{\sqrt{3}} - \frac{y}{\frac{1}{3}} \cdot \frac{1}{\sqrt{3}} = 1 \]
This simplifies to:
\[ \frac{2x}{\frac{1}{2}} \cdot \frac{1}{\sqrt{3}} - \frac{3y}{\frac{1}{3}} \cdot \frac{1}{\sqrt{3}} = 1 \]
\[ 4x \cdot \frac{1}{\sqrt{3}} - 3y \cdot \frac{1}{\sqrt{3}} = 1 \]

### Step 4: Clear the fraction
Multiplying through by \(\sqrt{3}\) to eliminate the denominator gives:
\[ 4x - 3y = \sqrt{3} \]

### Final Answer
Thus, the equation of the tangent to the hyperbola at the eccentric angle \(\frac{\pi}{6}\) is:
\[ 4x - 3y = \sqrt{3} \]