The area (in square units) of the equilateral triangle formed by the tangent at ` (sqrt3,0) ` to the hyperbola ` x^(2) -3y^(2) =3 `with the pair of asymptotes of the heyperbola is

Options

A. `sqrt2`
B. `sqrt3`
C. `1/sqrt3`
D. `2sqrt3`

Answer

To solve the problem of finding the area of the equilateral triangle formed by the tangent at the point \((\sqrt{3}, 0)\) to the hyperbola \(x^2 - 3y^2 = 3\) and the pair of asymptotes of the hyperbola, we will follow these steps:

### Step 1: Write the equation of the hyperbola
The given hyperbola is:
\[ x^2 - 3y^2 = 3 \]
Dividing the entire equation by 3, we get:
\[ \frac{x^2}{3} - \frac{y^2}{1} = 1 \]
From this, we can identify \(a^2 = 3\) and \(b^2 = 1\).

### Step 2: Find the equation of the tangent at the point \((\sqrt{3}, 0)\)
The equation of the tangent to the hyperbola at the point \((x_1, y_1)\) is given by:
\[ \frac{xx_1}{a^2} - \frac{yy_1}{b^2} = 1 \]
Substituting \(x_1 = \sqrt{3}\) and \(y_1 = 0\):
\[ \frac{x \cdot \sqrt{3}}{3} - \frac{y \cdot 0}{1} = 1 \]
This simplifies to:
\[ \frac{\sqrt{3}}{3} x = 1 \quad \Rightarrow \quad x = \sqrt{3} \]

### Step 3: Write the equations of the asymptotes
The equations of the asymptotes for the hyperbola are given by:
\[ y = \pm \frac{b}{a} x \]
Substituting \(b = 1\) and \(a = \sqrt{3}\):
\[ y = \pm \frac{1}{\sqrt{3}} x \]

### Step 4: Find the points of intersection of the tangent and the asymptotes
1. For the asymptote \(y = \frac{1}{\sqrt{3}} x\):
\[ y = \frac{1}{\sqrt{3}} \cdot \sqrt{3} = 1 \quad \Rightarrow \quad P = (\sqrt{3}, 1) \]

2. For the asymptote \(y = -\frac{1}{\sqrt{3}} x\):
\[ y = -\frac{1}{\sqrt{3}} \cdot \sqrt{3} = -1 \quad \Rightarrow \quad Q = (\sqrt{3}, -1) \]

### Step 5: Identify the vertices of the triangle
The vertices of the triangle formed by the tangent and the asymptotes are:
- \(O(0, 0)\) (the origin)
- \(P(\sqrt{3}, 1)\)
- \(Q(\sqrt{3}, -1)\)

### Step 6: Calculate the area of triangle \(OPQ\)
The area \(A\) of triangle \(OPQ\) can be calculated using the formula:
\[ A = \frac{1}{2} \times \text{base} \times \text{height} \]
Here, the base \(PQ\) is the vertical distance between points \(P\) and \(Q\), which is:
\[ PQ = 1 - (-1) = 2 \]
The height from point \(O\) to line \(PQ\) is the horizontal distance from the y-axis to \(x = \sqrt{3}\), which is \(\sqrt{3}\).

Thus, the area is:
\[ A = \frac{1}{2} \times 2 \times \sqrt{3} = \sqrt{3} \]

### Final Answer
The area of the equilateral triangle formed by the tangent and the asymptotes is:
\[ \sqrt{3} \text{ square units} \]