Find the equation of an ellipse whose eccentricity is 2/3, the latus rectum is 5 and the centre is at the origin.

Explanation

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

### Step 1: Understand the parameters of the ellipse
We are given:
- Eccentricity \( e = \frac{2}{3} \)
- Length of the latus rectum \( L = 5 \)
- Center at the origin

Since the eccentricity is less than 1, we know this is an ellipse. We will assume the form of the ellipse as:

\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]

where \( a > b \).

### Step 2: Use the formula for the latus rectum
The length of the latus rectum \( L \) for an ellipse is given by:

\[ L = \frac{2b^2}{a} \]

Substituting the value of \( L \):

\[ 5 = \frac{2b^2}{a} \]

From this, we can express \( b^2 \) in terms of \( a \):

\[ b^2 = \frac{5a}{2} \]

### Step 3: Use the relationship between \( a \), \( b \), and \( e \)
The eccentricity \( e \) is related to \( a \) and \( b \) by the equation:

\[ e^2 = 1 - \frac{b^2}{a^2} \]

Substituting the value of \( e \):

\[ \left(\frac{2}{3}\right)^2 = 1 - \frac{b^2}{a^2} \]

Calculating \( e^2 \):

\[ \frac{4}{9} = 1 - \frac{b^2}{a^2} \]

Rearranging gives:

\[ \frac{b^2}{a^2} = 1 - \frac{4}{9} = \frac{5}{9} \]

### Step 4: Substitute \( b^2 \) into the equation
Now, we can substitute \( b^2 = \frac{5a}{2} \) into the equation:

\[ \frac{\frac{5a}{2}}{a^2} = \frac{5}{9} \]

This simplifies to:

\[ \frac{5}{2a} = \frac{5}{9} \]

### Step 5: Solve for \( a \)
Cross-multiplying gives:

\[ 5 \cdot 9 = 5 \cdot 2a \]

This simplifies to:

\[ 45 = 10a \implies a = \frac{45}{10} = \frac{9}{2} \]

### Step 6: Find \( b^2 \)
Now that we have \( a \), we can find \( b^2 \):

\[ b^2 = \frac{5a}{2} = \frac{5 \cdot \frac{9}{2}}{2} = \frac{45}{4} \]

### Step 7: Write the equation of the ellipse
Now we can substitute \( a^2 \) and \( b^2 \) into the equation of the ellipse:

\[ \frac{x^2}{\left(\frac{9}{2}\right)^2} + \frac{y^2}{\left(\frac{\sqrt{45}}{2}\right)^2} = 1 \]

This simplifies to:

\[ \frac{x^2}{\frac{81}{4}} + \frac{y^2}{\frac{45}{4}} = 1 \]

Multiplying through by 4 gives:

\[ \frac{4x^2}{81} + \frac{4y^2}{45} = 1 \]

### Final Equation
Thus, the equation of the ellipse is:

\[ \frac{4x^2}{81} + \frac{4y^2}{45} = 1 \]