If point (x,y) satisfy the relation `x^2+4y^2-4x+3=0` then

Options

A. `x in [1,3] , y in [-1/3,1/3]`
B. `x in [1,3] , y in [1,3]`
C. `x in [1,3] , y in [-1/2,1/2]`
D. `x in [-2,2] , y in [-1,1]`

Answer

To solve the equation \( x^2 + 4y^2 - 4x + 3 = 0 \) and determine the ranges for \( x \) and \( y \), we can follow these steps:

### Step 1: Rearranging the Equation
Start by rearranging the given equation:
\[ x^2 - 4x + 4y^2 + 3 = 0 \]

### Step 2: Completing the Square for \( x \)
Now, we will complete the square for the \( x \) terms:
\[ (x^2 - 4x + 4) + 4y^2 + 3 - 4 = 0 \]
This simplifies to:
\[ (x - 2)^2 + 4y^2 - 1 = 0 \]
Rearranging gives:
\[ (x - 2)^2 + 4y^2 = 1 \]

### Step 3: Recognizing the Shape
The equation \( (x - 2)^2 + 4y^2 = 1 \) represents an ellipse centered at \( (2, 0) \).

### Step 4: Identifying the Axes
From the standard form of the ellipse, we can identify:
- The semi-major axis \( a = 1 \) (along the \( x \)-axis)
- The semi-minor axis \( b = \frac{1}{2} \) (along the \( y \)-axis)

### Step 5: Finding the Range for \( x \)
The range for \( x \) can be determined from the center and the semi-major axis:
\[ x - 2 \in [-1, 1] \implies 1 \leq x \leq 3 \]

### Step 6: Finding the Range for \( y \)
The range for \( y \) can be determined from the semi-minor axis:
\[ y \in \left[-\frac{1}{2}, \frac{1}{2}\right] \]

### Conclusion
Thus, the ranges for \( x \) and \( y \) are:
- \( x \) belongs to \( [1, 3] \)
- \( y \) belongs to \( \left[-\frac{1}{2}, \frac{1}{2}\right] \)