If parametric representation of a parabola is `x=2+t^(2) and y=2t+1`, then

Options

A. axis of parabola is y=1
B. equation of directrix is x=1
C. focus of parabola is S(3,1)
D. vertex of parabola is V(2,1)

Answer

To solve the problem, we need to analyze the given parametric equations of the parabola, which are:

\[ x = 2 + t^2 \]
\[ y = 2t + 1 \]

### Step 1: Eliminate the parameter \( t \)

From the equation for \( y \), we can express \( t \) in terms of \( y \):

\[ y - 1 = 2t \implies t = \frac{y - 1}{2} \]

Now, substitute \( t \) into the equation for \( x \):

\[ x = 2 + t^2 = 2 + \left(\frac{y - 1}{2}\right)^2 \]

### Step 2: Simplify the equation

Now, we simplify the equation for \( x \):

\[ x = 2 + \frac{(y - 1)^2}{4} \]

Rearranging this gives:

\[ (y - 1)^2 = 4(x - 2) \]

### Step 3: Identify the vertex

The equation \((y - 1)^2 = 4(x - 2)\) is in the standard form of a parabola, which is \((y - k)^2 = 4a(x - h)\). Here, we can identify:

- Vertex \((h, k) = (2, 1)\)
- The value of \( 4a = 4 \implies a = 1\)

### Step 4: Determine the axis of the parabola

The axis of the parabola is horizontal since it opens to the right. The equation of the axis is given by the line \(y = k\):

\[ y = 1 \]

### Step 5: Find the focus

The focus of the parabola is located at a distance \(a\) from the vertex along the axis of symmetry. Since \(a = 1\) and the parabola opens to the right, the focus is at:

\[ (h + a, k) = (2 + 1, 1) = (3, 1) \]

### Step 6: Find the directrix

The directrix is a vertical line located \(a\) units to the left of the vertex. Therefore, the equation of the directrix is:

\[ x = h - a = 2 - 1 = 1 \]

### Summary of Results

- **Vertex**: \((2, 1)\)
- **Focus**: \((3, 1)\)
- **Directrix**: \(x = 1\)
- **Axis**: \(y = 1\)