The focus of the conic represented parametrically by the equation `y=t^(2)+3, x= 2t-1`is

Answer

To find the focus of the conic represented parametrically by the equations \( y = t^2 + 3 \) and \( x = 2t - 1 \), we can follow these steps:

### Step 1: Express \( t^2 \) in terms of \( y \)
From the equation \( y = t^2 + 3 \), we can isolate \( t^2 \):
\[ t^2 = y - 3 \quad \text{(Equation 1)} \]

### Step 2: Express \( t \) in terms of \( x \)
From the equation \( x = 2t - 1 \), we can isolate \( t \):
\[ 2t = x + 1 \implies t = \frac{x + 1}{2} \quad \text{(Equation 2)} \]

### Step 3: Substitute \( t \) into \( t^2 \)
Now, we substitute Equation 2 into Equation 1 to express \( y \) in terms of \( x \):
\[ t^2 = \left(\frac{x + 1}{2}\right)^2 = \frac{(x + 1)^2}{4} \]
Setting this equal to Equation 1:
\[ y - 3 = \frac{(x + 1)^2}{4} \]

### Step 4: Rearrange to standard form
Rearranging gives:
\[ 4(y - 3) = (x + 1)^2 \]
This can be rewritten as:
\[ (x + 1)^2 = 4(y - 3) \]

### Step 5: Identify the conic type
This equation is in the standard form of a parabola:
\[ (x - h)^2 = 4p(y - k) \]
where \( (h, k) \) is the vertex and \( p \) is the distance from the vertex to the focus. Here, \( h = -1 \), \( k = 3 \), and \( 4p = 4 \) implies \( p = 1 \).

### Step 6: Determine the focus
For a parabola that opens upward, the focus is located at:
\[ (h, k + p) = (-1, 3 + 1) = (-1, 4) \]

Thus, the focus of the conic is:
\[ \boxed{(-1, 4)} \]