The parametric coordinates of any point on the parabola whose focus is (0,1) and the directrix is `x+2=0`, are :

Options

A. `(t^2+1,2t-1)`
B. `(t^2+1,2t+1)`
C. `(t^2,2t)`
D. `(t^2-1,2t+1)`

Answer

To find the parametric coordinates of any point on the parabola whose focus is at (0, 1) and the directrix is given by the equation \(x + 2 = 0\) (or \(x = -2\)), we can follow these steps:

### Step 1: Identify the Focus and Directrix
The focus of the parabola is at the point \(F(0, 1)\) and the directrix is the line \(x = -2\).

### Step 2: Determine the Vertex
The vertex of the parabola lies midway between the focus and the directrix. The x-coordinate of the vertex can be found by averaging the x-coordinates of the focus and the directrix.

- Focus x-coordinate: \(0\)
- Directrix x-coordinate: \(-2\)

Vertex x-coordinate:
\[ x_v = \frac{0 + (-2)}{2} = -1 \]

The y-coordinate of the vertex is the same as that of the focus since the directrix is vertical:
\[ y_v = 1 \]

Thus, the vertex \(V\) is at \((-1, 1)\).

### Step 3: Determine the Distance \(a\)
The distance \(a\) from the vertex to the focus (or from the vertex to the directrix) is:
\[ a = |0 - (-1)| = 1 \]

### Step 4: Write the Standard Form of the Parabola
Since the parabola opens to the right (the focus is to the right of the directrix), the standard form of the parabola is:
\[ (y - k)^2 = 4a(x - h) \]
where \((h, k)\) is the vertex. Substituting \(h = -1\), \(k = 1\), and \(a = 1\):
\[ (y - 1)^2 = 4 \cdot 1 \cdot (x + 1) \]
This simplifies to:
\[ (y - 1)^2 = 4(x + 1) \]

### Step 5: Parametrize the Parabola
To find the parametric equations, we can let:
\[ x + 1 = t^2 \quad \text{(where \(t\) is a parameter)} \]
This gives:
\[ x = t^2 - 1 \]

Next, we can find \(y\) using the equation of the parabola:
\[ y - 1 = 2t \quad \Rightarrow \quad y = 2t + 1 \]

### Final Parametric Coordinates
Thus, the parametric coordinates of any point on the parabola are:
\[ (x, y) = (t^2 - 1, 2t + 1) \]

### Summary
The parametric coordinates of any point on the parabola are given by:
\[ (x, y) = (t^2 - 1, 2t + 1) \]