Let $ L_1,\ L_2$ be the ray with the origin as end point in a plane . $ L_1$ makes angle of $ \minus{}\alpha$ with the $ x$ axis and $ L_2$ makes angle of $ \alpha$ with the $ x$ axis, where $ 0<\alpha <\frac{\pi}{3}$. Take points $ P,\ Q$ on $ L_1,\ L_2$ respectively such that $ PQ\equal{}1$ and $ R$ in the opposite side  to $ O$ with respect to the line $ PQ$ such that $ \triangle{PQR}$ is regular triangle.

(1) If the segment $ PQ$ is perpendicular to the $ x$ axis, then find the coordinate of $ R$.

(2) When two points $ P,\ Q$ move on $ L_1,\ L_2$ with keeping $ PQ\equal{}1$, prove that the locus of $ R$ is a part of some ellipse.
### Part 1: Finding the coordinates of \( R \)

1. **Identify the coordinates of \( P \) and \( Q \)**:
   - Since \( L_1 \) makes an angle of \( -\alpha \) with the \( x \)-axis, any point \( P \) on \( L_1 \) can be written as \( P = (r \cos(-\alpha), r \sin(-\alpha)) \).
   - Similarly, since \( L_2 \) makes an angle of \( \alpha \) with the \( x \)-axis, any point \( Q \) on \( L_2 \) can be written as \( Q = (s \cos(\alpha), s \sin(\alpha)) \).

2. **Given \( PQ = 1 \) and \( PQ \) is perpendicular to the \( x \)-axis**:
   - This implies that the \( y \)-coordinates of \( P \) and \( Q \) must be the same, i.e., \( r \sin(-\alpha) = s \sin(\alpha) \).
   - Since \( \sin(-\alpha) = -\sin(\alpha) \), we have \( r (-\sin(\alpha)) = s \sin(\alpha) \), which simplifies to \( r = -s \).

3. **Calculate the coordinates of \( P \) and \( Q \)**:
   - Let \( P = (r \cos(-\alpha), r \sin(-\alpha)) = (r \cos(\alpha), -r \sin(\alpha)) \).
   - Let \( Q = (s \cos(\alpha), s \sin(\alpha)) = (-r \cos(\alpha), r \sin(\alpha)) \).

4. **Find the midpoint \( M \) of \( PQ \)**:
   - The midpoint \( M \) of \( PQ \) is given by:
     \[
     M = \left( \frac{r \cos(\alpha) + (-r \cos(\alpha))}{2}, \frac{-r \sin(\alpha) + r \sin(\alpha)}{2} \right) = (0, 0)
     \]

5. **Determine the coordinates of \( R \)**:
   - Since \( \triangle PQR \) is a regular triangle, \( R \) is obtained by rotating \( P \) by \( 60^\circ \) counterclockwise around \( Q \).
   - The rotation matrix for \( 60^\circ \) is:
     \[
     \begin{pmatrix}
     \cos(60^\circ) & -\sin(60^\circ) \\
     \sin(60^\circ) & \cos(60^\circ)
     \end{pmatrix}
     = \begin{pmatrix}
     \frac{1}{2} & -\frac{\sqrt{3}}{2} \\
     \frac{\sqrt{3}}{2} & \frac{1}{2}
     \end{pmatrix}
     \]
   - Applying this rotation to \( P \) with respect to \( Q \):
     \[
     R = Q + \begin{pmatrix}
     \frac{1}{2} & -\frac{\sqrt{3}}{2} \\
     \frac{\sqrt{3}}{2} & \frac{1}{2}
     \end{pmatrix} \begin{pmatrix}
     P_x - Q_x \\
     P_y - Q_y
     \end{pmatrix}
     \]
   - Substituting \( P = (r \cos(\alpha), -r \sin(\alpha)) \) and \( Q = (-r \cos(\alpha), r \sin(\alpha)) \):
     \[
     R = \left( -r \cos(\alpha), r \sin(\alpha) \right) + \begin{pmatrix}
     \frac{1}{2} & -\frac{\sqrt{3}}{2} \\
     \frac{\sqrt{3}}{2} & \frac{1}{2}
     \end{pmatrix} \begin{pmatrix}
     2r \cos(\alpha) \\
     -2r \sin(\alpha)
     \end{pmatrix}
     \]
   - Simplifying the matrix multiplication:
     \[
     R = \left( -r \cos(\alpha), r \sin(\alpha) \right) + \begin{pmatrix}
     r \cos(\alpha) - \sqrt{3} r \sin(\alpha) \\
     \sqrt{3} r \cos(\alpha) + r \sin(\alpha)
     \end{pmatrix}
     \]
     \[
     R = \left( -r \cos(\alpha) + r \cos(\alpha) - \sqrt{3} r \sin(\alpha), r \sin(\alpha) + \sqrt{3} r \cos(\alpha) + r \sin(\alpha) \right)
     \]
     \[
     R = \left( -\sqrt{3} r \sin(\alpha), (\sqrt{3} \cos(\alpha) + 2 \sin(\alpha)) r \right)
     \]

### Part 2: Proving the locus of \( R \) is part of an ellipse

1. **Express the coordinates of \( R \) in terms of \( \theta \)**:
   - Let \( \theta \) be the angle formed by line \( L_1 \) and \( PQ \).
   - Using trigonometry, we can express the coordinates of \( R \) as:
     \[
     R_x = \sqrt{3} \sin(\theta + \frac{\pi}{6})
     \]
     \[
     R_y = \frac{1}{\sqrt{3}} \cos(\theta + \frac{\pi}{6})
     \]

2. **Find the equation of the locus**:
   - Substitute \( R_x \) and \( R_y \) into the equation of an ellipse:
     \[
     \frac{R_x^2}{3} + 3 R_y^2 = 1
     \]
   - Simplifying, we get:
     \[
     \frac{(\sqrt{3} \sin(\theta + \frac{\pi}{6}))^2}{3} + 3 \left( \frac{\cos(\theta + \frac{\pi}{6})}{\sqrt{3}} \right)^2 = 1
     \]
     \[
     \sin^2(\theta + \frac{\pi}{6}) + \cos^2(\theta + \frac{\pi}{6}) = 1
     \]
   - This is always true, confirming that the locus of \( R \) is part of an ellipse.

The final answer is \(\boxed{\left( -\sqrt{3} r \sin(\alpha), (\sqrt{3} \cos(\alpha) + 2 \sin(\alpha)) r \right)}\)