A tangent drawn to the hyperbola `(x^2)/(a^2) - (y^2)/(b^2) = 1` at `P(a "sec" pi/6, b tan pi/6)` form a triangle of area `3a^2` sq. units with the coordinate axes. The eccentricity of the conjugate hyperbola of `(x^2)/(a^2) - (y^2)/(b^2) = 1` is

Options

A. `sqrt(17)`
B. `(sqrt17)/(4)`
C. `(sqrt17)/(2)`
D. `8/(sqrt17)`

Answer

To solve the problem, we will follow these steps:

### Step 1: Identify the point on the hyperbola
The point \( P \) on the hyperbola is given as:
\[ P\left(a \sec \frac{\pi}{6}, b \tan \frac{\pi}{6}\right) \]
Calculating the trigonometric values:
\[ \sec \frac{\pi}{6} = \frac{2}{\sqrt{3}}, \quad \tan \frac{\pi}{6} = \frac{1}{\sqrt{3}} \]
Thus, the coordinates of point \( P \) are:
\[ P\left(a \cdot \frac{2}{\sqrt{3}}, b \cdot \frac{1}{\sqrt{3}}\right) = \left(\frac{2a}{\sqrt{3}}, \frac{b}{\sqrt{3}}\right) \]

### Step 2: Write the equation of the tangent at point \( P \)
The equation of the tangent to the hyperbola \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) at point \( (x_1, y_1) \) 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:
\[ \frac{2x}{a\sqrt{3}} - \frac{y}{b\sqrt{3}} = 1 \]

### Step 3: Convert to intercept form
Multiplying through by \( a b \sqrt{3} \):
\[ 2bx - ay = ab\sqrt{3} \]
Rearranging gives:
\[ 2bx - ay = ab\sqrt{3} \]
This can be expressed in intercept form:
\[ \frac{x}{\frac{ab\sqrt{3}}{2b}} + \frac{y}{\frac{ab\sqrt{3}}{a}} = 1 \]
Thus, the x-intercept is \( \frac{ab\sqrt{3}}{2b} = \frac{a\sqrt{3}}{2} \) and the y-intercept is \( \frac{ab\sqrt{3}}{a} = b\sqrt{3} \).

### Step 4: Calculate the area of the triangle formed
The area \( A \) of the triangle formed by the intercepts with the axes is given by:
\[ A = \frac{1}{2} \times \text{x-intercept} \times \text{y-intercept} = \frac{1}{2} \times \frac{a\sqrt{3}}{2} \times b\sqrt{3} \]
This simplifies to:
\[ A = \frac{3ab}{4} \]

### Step 5: Set the area equal to \( 3a^2 \)
According to the problem, this area is given as \( 3a^2 \):
\[ \frac{3ab}{4} = 3a^2 \]
Cancelling \( 3 \) from both sides gives:
\[ \frac{ab}{4} = a^2 \]
Thus, multiplying both sides by \( 4 \):
\[ ab = 4a^2 \]
Dividing both sides by \( a \) (assuming \( a \neq 0 \)):
\[ b = 4a \]

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

### Final Answer
The eccentricity of the conjugate hyperbola is:
\[ \frac{\sqrt{17}}{4} \]