The angle between the lines joining the origin to the points of intersection of the line `sqrt3x+y=2` and the curve `y^(2)-x^(2)=4` is

Options

A. `tan^(-1)(2//sqrt3)`
B. `pi//6`
C. `tan^(-1)(sqrt3//2)`
D. `pi//2`

Explanation

To find the angle between the lines joining the origin to the points of intersection of the line \(\sqrt{3}x + y = 2\) and the curve \(y^2 - x^2 = 4\), we will follow these steps:

### Step 1: Find the points of intersection
We need to solve the equations simultaneously. Start with the line equation:
\[ y = 2 - \sqrt{3}x \]
Substituting this expression for \(y\) into the curve equation \(y^2 - x^2 = 4\):
\[ (2 - \sqrt{3}x)^2 - x^2 = 4 \]

### Step 2: Expand and simplify
Expanding the left-hand side:
\[ (2 - \sqrt{3}x)(2 - \sqrt{3}x) - x^2 = 4 \]
\[ 4 - 4\sqrt{3}x + 3x^2 - x^2 = 4 \]
\[ 2x^2 - 4\sqrt{3}x + 4 - 4 = 0 \]
\[ 2x^2 - 4\sqrt{3}x = 0 \]

### Step 3: Factor the equation
Factoring out \(2x\):
\[ 2x(x - 2\sqrt{3}) = 0 \]
This gives us two solutions:
1. \(x = 0\)
2. \(x = 2\sqrt{3}\)

### Step 4: Find corresponding \(y\) values
For \(x = 0\):
\[ y = 2 - \sqrt{3}(0) = 2 \quad \Rightarrow \quad (0, 2) \]
For \(x = 2\sqrt{3}\):
\[ y = 2 - \sqrt{3}(2\sqrt{3}) = 2 - 6 = -4 \quad \Rightarrow \quad (2\sqrt{3}, -4) \]

### Step 5: Find slopes of the lines
Now we have two points of intersection: \((0, 2)\) and \((2\sqrt{3}, -4)\). We will find the slopes of the lines from the origin to these points.

1. Slope to \((0, 2)\):
\[ m_1 = \frac{2 - 0}{0 - 0} \quad \text{(undefined, vertical line)} \]

2. Slope to \((2\sqrt{3}, -4)\):
\[ m_2 = \frac{-4 - 0}{2\sqrt{3} - 0} = \frac{-4}{2\sqrt{3}} = -\frac{2}{\sqrt{3}} \]

### Step 6: Find the angle between the lines
The angle \(\theta\) between two lines with slopes \(m_1\) and \(m_2\) can be found using the formula:
\[ \tan(\theta) = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| \]
Since \(m_1\) is undefined (vertical line), we can directly calculate the angle between the vertical line and the line with slope \(m_2\).

The angle \(\theta\) between a vertical line and a line with slope \(m_2\) is given by:
\[ \theta = \tan^{-1}\left(-\frac{2}{\sqrt{3}}\right) \]

### Step 7: Calculate the angle
To find the angle between the two lines:
\[ \theta = 90^\circ - \tan^{-1}\left(\frac{2}{\sqrt{3}}\right) \]
This can also be expressed as:
\[ \theta = \tan^{-1}\left(\frac{\sqrt{3}}{2}\right) \]

### Final Answer
Thus, the angle between the lines joining the origin to the points of intersection is:
\[ \theta = \tan^{-1}\left(\frac{\sqrt{3}}{2}\right) \]