A particle moves in the x-y plane. It x and y coordinates vary with time t according to equations `x=t^(2)+2t` and `y=2t`. Possible shape of path followed by the particle is

Options

A. Straight line
B. Circle
C. Parabola
D. More information Is required to decide.

Answer

To determine the shape of the path followed by the particle moving in the x-y plane, we start with the given equations for x and y:

1. **Equations**:
- \( x = t^2 + 2t \)
- \( y = 2t \)

2. **Express \( t \) in terms of \( y \)**:
From the equation for \( y \):
\[ y = 2t \implies t = \frac{y}{2} \]

3. **Substitute \( t \) into the equation for \( x \)**:
Now we substitute \( t \) into the equation for \( x \):
\[ x = \left(\frac{y}{2}\right)^2 + 2\left(\frac{y}{2}\right) \]
Simplifying this:
\[ x = \frac{y^2}{4} + y \]

4. **Rearranging the equation**:
We can rearrange this equation to express it in a standard form:
\[ x = \frac{y^2}{4} + y \implies 4x = y^2 + 4y \]
Rearranging gives:
\[ y^2 + 4y - 4x = 0 \]

5. **Completing the square**:
To complete the square for the \( y \) terms:
\[ y^2 + 4y = (y + 2)^2 - 4 \]
Thus, we rewrite the equation:
\[ (y + 2)^2 - 4 = 4x \implies (y + 2)^2 = 4x + 4 \]
This can be rewritten as:
\[ (y + 2)^2 = 4(x + 1) \]

6. **Identifying the shape**:
The equation \( (y + 2)^2 = 4(x + 1) \) is in the standard form of a parabola that opens to the right. The vertex of this parabola is at the point (-1, -2).

7. **Conclusion**:
Therefore, the possible shape of the path followed by the particle is a **parabola**.