The focus and corresponding directrix of an ellipse are (3, 4) and `x+y-1=0` respectively. If the eccentricity of the ellipse is `(1)/(2)`, then the coordinates of the centre of the ellipse are

Options

A. (2, 3)
B. (4, 5)
C. (8, 9)
D. (1, 2)

Answer

To find the coordinates of the center of the ellipse given the focus and directrix, we can follow these steps:

### Step 1: Identify the given information
- Focus \( F(3, 4) \)
- Directrix \( x + y - 1 = 0 \)
- Eccentricity \( e = \frac{1}{2} \)

### Step 2: Find the distance from the focus to the directrix
The distance \( d \) from a point \( (x_1, y_1) \) to the line \( ax + by + c = 0 \) is given by the formula:
\[ d = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}} \]
For the directrix \( x + y - 1 = 0 \), we have \( a = 1, b = 1, c = -1 \). Plugging in the coordinates of the focus \( (3, 4) \):
\[ d = \frac{|1(3) + 1(4) - 1|}{\sqrt{1^2 + 1^2}} = \frac{|3 + 4 - 1|}{\sqrt{2}} = \frac{6}{\sqrt{2}} = 3\sqrt{2} \]

### Step 3: Relate the distance to the eccentricity
For an ellipse, the distance from the focus to the directrix is given by:
\[ d = \frac{a}{e} \]
where \( a \) is the semi-major axis. Substituting \( e = \frac{1}{2} \):
\[ 3\sqrt{2} = \frac{a}{\frac{1}{2}} \implies a = 3\sqrt{2} \cdot \frac{1}{2} = \frac{3\sqrt{2}}{2} \]

### Step 4: Find the distance from the focus to the center
The distance from the focus to the center \( C \) of the ellipse is given by:
\[ d_{FC} = ae \]
Substituting the values we found:
\[ d_{FC} = \frac{3\sqrt{2}}{2} \cdot \frac{1}{2} = \frac{3\sqrt{2}}{4} \]

### Step 5: Determine the direction of the center from the focus
The slope of the directrix \( x + y - 1 = 0 \) is \( -1 \), which means the angle \( \theta \) with respect to the x-axis is \( 45^\circ \) or \( \frac{\pi}{4} \). Therefore, the direction vector can be represented as:
\[ \left( \cos \frac{\pi}{4}, \sin \frac{\pi}{4} \right) = \left( \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right) \]

### Step 6: Calculate the coordinates of the center
The coordinates of the center \( C \) can be found using:
\[ C = F + d_{FC} \cdot \text{direction vector} \]
Substituting the values:
\[ C = (3, 4) + \frac{3\sqrt{2}}{4} \left( \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right) = \left( 3 + \frac{3}{4}, 4 + \frac{3}{4} \right) = \left( \frac{12}{4} + \frac{3}{4}, \frac{16}{4} + \frac{3}{4} \right) = \left( \frac{15}{4}, \frac{19}{4} \right) \]

### Final Answer
The coordinates of the center of the ellipse are:
\[ \left( \frac{15}{4}, \frac{19}{4} \right) \]