A tangent drawn to hyperbola `(x^(2))/(a^(2)) - (y^(2))/(b^(2))=1` at `P((pi)/(6))` forms a triangle of area `3a^(2)` square units, with coordinate axes. If the eccentricity of hyperbola is e, then the value of `e^(2)-9` is

Options

A. 9
B. 10
C. 11
D. 8

Answer

To solve the problem step by step, we will follow the given information and derive the required values systematically.

### Step 1: Identify the point on the hyperbola
The hyperbola is given by the equation:
\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \]
The point \( P \) is given as \( P\left(\frac{\pi}{6}\right) \). We can express the coordinates of point \( P \) in terms of \( a \) and \( b \):
\[ P\left(a \sec\left(\frac{\pi}{6}\right), b \tan\left(\frac{\pi}{6}\right)\right) \]
Calculating the trigonometric values:
\[ \sec\left(\frac{\pi}{6}\right) = \frac{2}{\sqrt{3}}, \quad \tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}} \]
Thus, the coordinates of point \( P \) become:
\[ P\left(a \cdot \frac{2}{\sqrt{3}}, b \cdot \frac{1}{\sqrt{3}}\right) = \left(\frac{2a}{\sqrt{3}}, \frac{b}{\sqrt{3}}\right) \]

### Step 2: Write the equation of the tangent
The equation of the tangent to the hyperbola at point \( P(x_1, y_1) \) is given by:
\[ \frac{xx_1}{a^2} - \frac{yy_1}{b^2} = 1 \]
Substituting \( x_1 = \frac{2a}{\sqrt{3}} \) and \( y_1 = \frac{b}{\sqrt{3}} \):
\[ \frac{x \cdot \frac{2a}{\sqrt{3}}}{a^2} - \frac{y \cdot \frac{b}{\sqrt{3}}}{b^2} = 1 \]
Simplifying this gives:
\[ \frac{2x}{a\sqrt{3}} - \frac{y}{b\sqrt{3}} = 1 \]
Multiplying through by \( a b \sqrt{3} \):
\[ 2bx - ay = ab\sqrt{3} \]

### Step 3: Find the intercepts
From the equation \( 2bx - ay = ab\sqrt{3} \), we can find the x-intercept and y-intercept:
- **X-intercept**: Set \( y = 0 \):
\[ 2bx = ab\sqrt{3} \implies x = \frac{ab\sqrt{3}}{2b} = \frac{a\sqrt{3}}{2} \]
- **Y-intercept**: Set \( x = 0 \):
\[ -ay = ab\sqrt{3} \implies y = -\frac{ab\sqrt{3}}{a} = -b\sqrt{3} \]

### Step 4: Area of the triangle formed by intercepts
The area \( A \) of the triangle formed by the intercepts on the axes is given by:
\[ A = \frac{1}{2} \times \text{(x-intercept)} \times \text{(y-intercept)} = \frac{1}{2} \times \frac{a\sqrt{3}}{2} \times (-b\sqrt{3}) \]
Calculating this gives:
\[ A = \frac{1}{2} \times \frac{a\sqrt{3}}{2} \times b\sqrt{3} = \frac{3ab}{4} \]

### Step 5: Set the area equal to given area
We are given that the area is \( 3a^2 \):
\[ \frac{3ab}{4} = 3a^2 \]
Dividing both sides by 3:
\[ \frac{ab}{4} = a^2 \implies ab = 4a^2 \implies b = 4a \quad (\text{assuming } a \neq 0) \]

### Step 6: Find the eccentricity
The eccentricity \( e \) of the hyperbola is given by:
\[ e^2 = 1 + \frac{b^2}{a^2} \]
Substituting \( b = 4a \):
\[ e^2 = 1 + \frac{(4a)^2}{a^2} = 1 + 16 = 17 \]

### Step 7: Calculate \( e^2 - 9 \)
Finally, we compute:
\[ e^2 - 9 = 17 - 9 = 8 \]

Thus, the final answer is:
\[ \boxed{8} \]