Find the equation of the ellipse whose centre is the origin, major axis `9/2` and eccentricity `1/sqrt(3)` where the major axis is the horizontal axis.

Answer

To find the equation of the ellipse with the given parameters, we will follow these steps:

### Step 1: Identify the parameters of the ellipse
The center of the ellipse is at the origin (0, 0). The length of the major axis is given as \( \frac{9}{2} \), and the eccentricity \( e \) is given as \( \frac{1}{\sqrt{3}} \).

### Step 2: Determine the semi-major axis \( a \)
The length of the major axis is \( 2a \), so we can find \( a \) as follows:
\[ 2a = \frac{9}{2} \implies a = \frac{9}{4} \]

### Step 3: Use the eccentricity to find the semi-minor axis \( b \)
The relationship between the semi-major axis \( a \), semi-minor axis \( b \), and eccentricity \( e \) is given by the formula:
\[ e = \frac{c}{a} \]
where \( c = \sqrt{a^2 - b^2} \).

From the eccentricity given:
\[ e = \frac{1}{\sqrt{3}} \implies c = a \cdot e = \frac{9}{4} \cdot \frac{1}{\sqrt{3}} = \frac{9}{4\sqrt{3}} \]

Now, we can find \( b \):
\[ c^2 = a^2 - b^2 \implies \left(\frac{9}{4\sqrt{3}}\right)^2 = \left(\frac{9}{4}\right)^2 - b^2 \]
Calculating \( c^2 \) and \( a^2 \):
\[ c^2 = \frac{81}{48} \quad \text{and} \quad a^2 = \frac{81}{16} \]
Substituting these into the equation:
\[ \frac{81}{48} = \frac{81}{16} - b^2 \]
Finding a common denominator (48):
\[ \frac{81}{48} = \frac{243}{48} - b^2 \implies b^2 = \frac{243}{48} - \frac{81}{48} = \frac{162}{48} = \frac{27}{8} \]

### Step 4: Write the equation of the ellipse
The standard form of the equation of an ellipse centered at the origin with a horizontal major axis is given by:
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]
Substituting \( a^2 = \frac{81}{16} \) and \( b^2 = \frac{27}{8} \):
\[ \frac{x^2}{\frac{81}{16}} + \frac{y^2}{\frac{27}{8}} = 1 \]

### Step 5: Simplifying the equation
To eliminate the fractions, we can multiply through by the least common multiple of the denominators, which is \( 16 \cdot 8 = 128 \):
\[ 128 \left( \frac{x^2}{\frac{81}{16}} + \frac{y^2}{\frac{27}{8}} \right) = 128 \cdot 1 \]
This simplifies to:
\[ \frac{128 \cdot 16}{81} x^2 + \frac{128 \cdot 8}{27} y^2 = 128 \]
This results in:
\[ \frac{16x^2}{81} + \frac{8y^2}{27} = 1 \]

### Final Equation
Thus, the equation of the ellipse is:
\[ \frac{16x^2}{81} + \frac{8y^2}{27} = 1 \]