Find the locus of a point whose coordinate are given by `x = t+t^(2), y = 2t+1`, where t is variable.

Answer

To find the locus of the point whose coordinates are given by \( x = t + t^2 \) and \( y = 2t + 1 \), we will eliminate the parameter \( t \) from these equations.

### Step-by-Step Solution:

1. **Express \( t \) in terms of \( y \)**:
Given the equation for \( y \):
\[ y = 2t + 1 \]
We can solve for \( t \):
\[ t = \frac{y - 1}{2} \]

**Hint**: Rearranging the equation for \( y \) will help isolate \( t \).

2. **Substitute \( t \) into the equation for \( x \)**:
Now that we have \( t \) in terms of \( y \), we can substitute this expression into the equation for \( x \):
\[ x = t + t^2 \]
Substituting \( t \):
\[ x = \frac{y - 1}{2} + \left(\frac{y - 1}{2}\right)^2 \]

**Hint**: Make sure to square the term correctly when substituting.

3. **Simplify the expression**:
First, calculate \( \left(\frac{y - 1}{2}\right)^2 \):
\[ \left(\frac{y - 1}{2}\right)^2 = \frac{(y - 1)^2}{4} = \frac{y^2 - 2y + 1}{4} \]
Now substitute this back into the equation for \( x \):
\[ x = \frac{y - 1}{2} + \frac{y^2 - 2y + 1}{4} \]
To combine these fractions, we need a common denominator:
\[ x = \frac{2(y - 1)}{4} + \frac{y^2 - 2y + 1}{4} \]
This simplifies to:
\[ x = \frac{2y - 2 + y^2 - 2y + 1}{4} = \frac{y^2 - 1}{4} \]

**Hint**: Ensure you combine like terms correctly when simplifying.

4. **Multiply through by 4**:
To eliminate the fraction, multiply both sides by 4:
\[ 4x = y^2 - 1 \]

**Hint**: Multiplying by a constant can help simplify the equation.

5. **Rearranging the equation**:
Rearranging gives us:
\[ y^2 = 4x + 1 \]

**Hint**: This is the standard form of a parabola.

### Final Result:
The locus of the point is given by the equation:
\[ y^2 = 4x + 1 \]

This represents a parabola that opens to the right.