The the vertex of the conic `y^(2)-4y=4x-4a` always lies between the straight lines `x+y=3 and 2x+2y-1=0` then

Options

A. `2 lt a lt 4`
B. `-(1)/(2) lt a lt 2`
C. `0 lt a lt 2`
D. `-(1)/(2) lt a lt (3)/(2)`

Answer

To solve the problem, we need to find the conditions under which the vertex of the conic \( y^2 - 4y = 4x - 4a \) lies between the lines \( x + y = 3 \) and \( 2x + 2y - 1 = 0 \).

### Step 1: Rewrite the conic equation
Starting with the equation of the conic:
\[ y^2 - 4y = 4x - 4a \]
We can rearrange it as:
\[ y^2 - 4y - 4x + 4a = 0 \]

### Step 2: Complete the square for \( y \)
To make it easier to find the vertex, we complete the square for the \( y \) terms:
\[ y^2 - 4y = (y - 2)^2 - 4 \]
Substituting this back into the equation gives:
\[ (y - 2)^2 - 4 = 4x - 4a \]
Rearranging, we have:
\[ (y - 2)^2 = 4x - 4a + 4 \]
\[ (y - 2)^2 = 4(x - a + 1) \]

### Step 3: Identify the vertex of the conic
From the equation \( (y - 2)^2 = 4(x - (a - 1)) \), we can see that the vertex of the conic is at the point:
\[ \text{Vertex} = (a - 1, 2) \]

### Step 4: Determine the conditions for the vertex to lie between the lines
We need to find the conditions under which the vertex \( (a - 1, 2) \) lies between the lines \( x + y = 3 \) and \( 2x + 2y - 1 = 0 \).

1. **Line 1: \( x + y = 3 \)**
- Substitute the vertex into the line equation:
\[ (a - 1) + 2 < 3 \implies a + 1 < 3 \implies a < 2 \]

2. **Line 2: \( 2x + 2y - 1 = 0 \)**
- Rearranging gives \( x + y = \frac{1}{2} \).
- Substitute the vertex into this line equation:
\[ 2(a - 1) + 2 < 1 \implies 2a - 2 < 1 \implies 2a < 3 \implies a < \frac{3}{2} \]

### Step 5: Combine the inequalities
From the two inequalities derived:
1. \( a < 2 \)
2. \( a < \frac{3}{2} \)

The more restrictive condition is:
\[ a < \frac{3}{2} \]

### Step 6: Find the lower bound for \( a \)
To ensure that the vertex lies between the lines, we also need to ensure that the vertex does not go below the line \( x + y = 3 \). Therefore, we need to check the lower bound:
\[ a - 1 + 2 > 0 \implies a + 1 > 0 \implies a > -1 \]

### Final Result
Combining both conditions, we find:
\[ -1 < a < \frac{3}{2} \]