The centre of a ellipse where axes is parllel to co-ordinate axes is (2, -1) and the semi axes are `sqrt(3)/(2),1/2` The equation of the ellipse is

Options

A. `12x^(2)+4y^(2)-16y+24x+25=0`
B. `12y^(2)+4x^(2)-16x+24y+25=0`
C. `12y^(2)+4x^(2)-16y+24yx+25=0`
D. `12x^(2)+4y^(2)-16y+24y+25=0`

Answer

To find the equation of the ellipse given its center and semi-axes, we can follow these steps:

### Step 1: Identify the center and semi-axes
The center of the ellipse is given as (h, k) = (2, -1). The semi-major axis \( a \) is given as \( \frac{\sqrt{3}}{2} \) and the semi-minor axis \( b \) is given as \( \frac{1}{2} \).

### Step 2: Write the standard form of the ellipse equation
The standard form of the equation of an ellipse centered at (h, k) with semi-major axis \( a \) and semi-minor axis \( b \) is:
\[ \frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1 \]

### Step 3: Substitute the values into the equation
Substituting \( h = 2 \), \( k = -1 \), \( a = \frac{\sqrt{3}}{2} \), and \( b = \frac{1}{2} \) into the equation:
\[ \frac{(x - 2)^2}{\left(\frac{\sqrt{3}}{2}\right)^2} + \frac{(y + 1)^2}{\left(\frac{1}{2}\right)^2} = 1 \]

### Step 4: Calculate \( a^2 \) and \( b^2 \)
Calculating \( a^2 \) and \( b^2 \):
\[ a^2 = \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{4} \]
\[ b^2 = \left(\frac{1}{2}\right)^2 = \frac{1}{4} \]

### Step 5: Substitute \( a^2 \) and \( b^2 \) back into the equation
Now substituting \( a^2 \) and \( b^2 \) into the equation:
\[ \frac{(x - 2)^2}{\frac{3}{4}} + \frac{(y + 1)^2}{\frac{1}{4}} = 1 \]

### Step 6: Clear the denominators
To eliminate the fractions, multiply the entire equation by 4:
\[ 4 \cdot \frac{(x - 2)^2}{\frac{3}{4}} + 4 \cdot \frac{(y + 1)^2}{\frac{1}{4}} = 4 \]
This simplifies to:
\[ \frac{4(x - 2)^2}{\frac{3}{4}} + 4(y + 1)^2 = 4 \]
\[ \frac{16(x - 2)^2}{3} + 4(y + 1)^2 = 4 \]

### Step 7: Rearranging the equation
Now, we can rearrange the equation:
\[ \frac{16(x - 2)^2}{3} + 4(y + 1)^2 - 4 = 0 \]

### Step 8: Final equation
To express it in a standard form, we can multiply through by 3 to eliminate the fraction:
\[ 16(x - 2)^2 + 12(y + 1)^2 - 12 = 0 \]
This can be rewritten as:
\[ 16(x - 2)^2 + 12(y + 1)^2 = 12 \]

### Conclusion
Thus, the equation of the ellipse is:
\[ \frac{16(x - 2)^2}{12} + \frac{12(y + 1)^2}{12} = 1 \]
or simplified:
\[ \frac{4(x - 2)^2}{3} + (y + 1)^2 = 1 \]