Find the equation to the ellipse with axes as the axes of coordinates.
latus rectum is 5 and eccentricity `2/3`,

Answer

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

### Step 1: Understand the given parameters
We are given:
- Latus rectum (L) = 5
- Eccentricity (e) = 2/3

### Step 2: Write the general equation of the ellipse
The standard form of the ellipse with axes along the coordinate axes is given by:
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]

### Step 3: Relate latus rectum to \(a\) and \(b\)
The latus rectum of an ellipse is given by the formula:
\[ L = \frac{2b^2}{a} \]
Substituting the value of the latus rectum:
\[ \frac{2b^2}{a} = 5 \]
From this, we can express \(b^2\) in terms of \(a\):
\[ b^2 = \frac{5a}{2} \quad \text{(Equation 1)} \]

### Step 4: Relate \(b^2\) to \(a^2\) and eccentricity
We also know that:
\[ b^2 = a^2(1 - e^2) \]
Substituting the value of eccentricity:
\[ b^2 = a^2\left(1 - \left(\frac{2}{3}\right)^2\right) = a^2\left(1 - \frac{4}{9}\right) = a^2\left(\frac{5}{9}\right) \]
Thus, we can express \(b^2\) in terms of \(a^2\):
\[ b^2 = \frac{5a^2}{9} \quad \text{(Equation 2)} \]

### Step 5: Equate the two expressions for \(b^2\)
From Equation 1 and Equation 2, we have:
\[ \frac{5a}{2} = \frac{5a^2}{9} \]
We can simplify this by multiplying both sides by 18 to eliminate the fractions:
\[ 45a = 10a^2 \]
Rearranging gives:
\[ 10a^2 - 45a = 0 \]
Factoring out \(5a\):
\[ 5a(2a - 9) = 0 \]
This gives us two solutions:
1. \(a = 0\) (not valid for an ellipse)
2. \(2a - 9 = 0 \Rightarrow a = \frac{9}{2}\)

### Step 6: Find \(b^2\) using \(a\)
Now substituting \(a = \frac{9}{2}\) back into Equation 1 to find \(b^2\):
\[ b^2 = \frac{5 \cdot \frac{9}{2}}{2} = \frac{45}{4} \]

### Step 7: Write the equation of the ellipse
Now we have \(a^2\) and \(b^2\):
- \(a^2 = \left(\frac{9}{2}\right)^2 = \frac{81}{4}\)
- \(b^2 = \frac{45}{4}\)

Thus, the equation of the ellipse is:
\[ \frac{x^2}{\frac{81}{4}} + \frac{y^2}{\frac{45}{4}} = 1 \]
Multiplying through by 4 gives:
\[ \frac{4x^2}{81} + \frac{4y^2}{45} = 1 \]

### Final Result
The equation of the ellipse is:
\[ \frac{4x^2}{81} + \frac{4y^2}{45} = 1 \]