If ` x^2 + 4y ^2 - 8x +12 =0` is satified by real values of x nad y then y must lies between

Options

A. 2, 6
B. 2, 5
C. -1, 1
D. -2, -1

Answer

To solve the equation \( x^2 + 4y^2 - 8x + 12 = 0 \) and determine the range of values for \( y \) such that the equation has real solutions for \( x \), we will follow these steps:

### Step 1: Rearranging the Equation
We start with the given equation:
\[ x^2 + 4y^2 - 8x + 12 = 0 \]
We can rearrange it to group the \( x \) terms:
\[ x^2 - 8x + 4y^2 + 12 = 0 \]

### Step 2: Completing the Square
Next, we complete the square for the \( x \) terms:
\[ x^2 - 8x = (x - 4)^2 - 16 \]
Substituting this back into the equation gives:
\[ (x - 4)^2 - 16 + 4y^2 + 12 = 0 \]
Simplifying this, we have:
\[ (x - 4)^2 + 4y^2 - 4 = 0 \]
Or,
\[ (x - 4)^2 + 4y^2 = 4 \]

### Step 3: Analyzing the Equation
This equation represents an ellipse centered at \( (4, 0) \) with semi-major axis 2 along the \( x \)-axis and semi-minor axis 1 along the \( y \)-axis. For \( (x - 4)^2 + 4y^2 = 4 \) to have real solutions, the term \( 4y^2 \) must be non-negative.

### Step 4: Setting Up the Inequality
To find the range of \( y \), we need to ensure that the expression remains non-negative:
\[ 4y^2 \leq 4 \]
Dividing through by 4 gives:
\[ y^2 \leq 1 \]

### Step 5: Solving the Inequality
Taking the square root of both sides, we find:
\[ -1 \leq y \leq 1 \]

### Conclusion
Thus, the values of \( y \) must lie in the interval:
\[ [-1, 1] \]

### Final Answer
The correct option from the choices provided is:
**Third option: -1 and 1**