The equation of the chord of the ellipse `x^(2) + 4y^(2) = 4` having the middle point at `(-2, (1)/(2))` is

Options

A. `2x - 2y + 7 = 0`
B. `x + 2y = 0`
C. `3x - 2y + 4 = 0`
D. `2x - 2y + 5 = 0`

Answer

To find the equation of the chord of the ellipse \( x^2 + 4y^2 = 4 \) with the midpoint at \( (-2, \frac{1}{2}) \), we can use the formula for the equation of a chord given its midpoint.

### Step-by-Step Solution:

1. **Identify the midpoint coordinates**:
The midpoint of the chord is given as \( (x_1, y_1) = (-2, \frac{1}{2}) \).

2. **Recall the formula for the chord**:
The equation of the chord of the ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) with midpoint \( (x_1, y_1) \) is given by:
\[ x \cdot x_1 + \frac{y \cdot y_1}{b^2} = \frac{a^2}{b^2} \cdot (x_1^2 + \frac{y_1^2}{b^2}) \]
For our ellipse \( x^2 + 4y^2 = 4 \), we can rewrite it in standard form:
\[ \frac{x^2}{4} + \frac{y^2}{1} = 1 \]
Here, \( a^2 = 4 \) and \( b^2 = 1 \).

3. **Substituting values into the formula**:
Substitute \( x_1 = -2 \) and \( y_1 = \frac{1}{2} \) into the chord equation:
\[ x \cdot (-2) + 4y \cdot \frac{1}{2} = 4 \cdot \left((-2)^2 + 4 \cdot \left(\frac{1}{2}\right)^2\right) \]

4. **Calculate the right side**:
\[ 4 \cdot (4 + 4 \cdot \frac{1}{4}) = 4 \cdot (4 + 1) = 4 \cdot 5 = 20 \]

5. **Form the equation**:
The left side becomes:
\[ -2x + 2y = 20 \]
Rearranging gives:
\[ 2x - 2y + 20 = 0 \]
Dividing through by 2:
\[ x - y + 10 = 0 \]

6. **Final equation of the chord**:
Thus, the equation of the chord is:
\[ x - y + 10 = 0 \]