A tangent drawn to hyperbola `x^2/a^2-y^2/b^2 = 1` at `P(pi/6)` froms a triangle of area `3a^2` square units, with the coordinate axes, then the square of its eccentricity is (A) `15` (B) `24` (C) `17` (D) `14`

Options

A. 15
B. 24
C. 17
D. 14

Explanation

To solve the problem, we need to find the square of the eccentricity of the hyperbola given the area of the triangle formed by the tangent at point \( P(\pi/6) \) and the coordinate axes.

### Step-by-Step Solution:

1. **Identify the Point on the Hyperbola:**
The hyperbola is given by the equation:
\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \]
For the angle \( \theta = \frac{\pi}{6} \), the coordinates of point \( P \) on the hyperbola can be expressed as:
\[ P = (a \sec(\frac{\pi}{6}), b \tan(\frac{\pi}{6})) \]
Using the values \( \sec(\frac{\pi}{6}) = \frac{2}{\sqrt{3}} \) and \( \tan(\frac{\pi}{6}) = \frac{1}{\sqrt{3}} \), we find:
\[ P = \left( a \cdot \frac{2}{\sqrt{3}}, b \cdot \frac{1}{\sqrt{3}} \right) = \left( \frac{2a}{\sqrt{3}}, \frac{b}{\sqrt{3}} \right) \]

2. **Equation of the Tangent:**
The equation of the tangent to the hyperbola at point \( P \) is given by:
\[ \frac{2x}{\sqrt{3}a} - \frac{y}{\sqrt{3}b} = 1 \]
Rearranging gives:
\[ 2x - \frac{a}{b}y = \sqrt{3}a \]

3. **Finding the Intercepts:**
To find the x-intercept, set \( y = 0 \):
\[ 2x = \sqrt{3}a \implies x = \frac{\sqrt{3}a}{2} \]
To find the y-intercept, set \( x = 0 \):
\[ -\frac{a}{b}y = \sqrt{3}a \implies y = -\frac{\sqrt{3}b}{1} \]

4. **Area of the Triangle:**
The area \( A \) of the triangle formed by the intercepts and the origin is given by:
\[ A = \frac{1}{2} \times \text{(base)} \times \text{(height)} = \frac{1}{2} \times \frac{\sqrt{3}a}{2} \times \sqrt{3}b \]
This simplifies to:
\[ A = \frac{3ab}{4} \]
We know from the problem statement that the area is \( 3a^2 \):
\[ \frac{3ab}{4} = 3a^2 \implies ab = 4a^2 \implies b = 4a \]

5. **Finding the Eccentricity:**
The eccentricity \( e \) of the hyperbola is given by:
\[ e^2 = 1 + \frac{b^2}{a^2} \]
Substituting \( b = 4a \):
\[ e^2 = 1 + \frac{(4a)^2}{a^2} = 1 + \frac{16a^2}{a^2} = 1 + 16 = 17 \]

### Final Answer:
The square of the eccentricity is:
\[ \boxed{17} \]