Obtain the equation of the ellipse whose latus rectum is 5 and whose e centricity is `(2)/(3)`, the axes of the ellipse being the axes of coordinates.

Answer

To find the equation of the ellipse with a given latus rectum and eccentricity, we can follow these steps:

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

### Step 2: Use the formula for the latus rectum of an ellipse
The latus rectum (L) of an ellipse is given by the formula:
\[ L = \frac{2b^2}{a} \]
where \( a \) is the semi-major axis and \( b \) is the semi-minor axis.

### Step 3: Substitute the known value of the latus rectum
Substituting the value of the latus rectum into the formula:
\[ 5 = \frac{2b^2}{a} \]
Rearranging gives:
\[ 2b^2 = 5a \quad \text{(1)} \]

### Step 4: Use the relationship between eccentricity, a, and b
The eccentricity (e) of an ellipse is defined as:
\[ e = \sqrt{1 - \frac{b^2}{a^2}} \]
Substituting the value of eccentricity:
\[ \frac{2}{3} = \sqrt{1 - \frac{b^2}{a^2}} \]
Squaring both sides:
\[ \left(\frac{2}{3}\right)^2 = 1 - \frac{b^2}{a^2} \]
This simplifies to:
\[ \frac{4}{9} = 1 - \frac{b^2}{a^2} \]
Rearranging gives:
\[ \frac{b^2}{a^2} = 1 - \frac{4}{9} = \frac{5}{9} \]
Thus, we have:
\[ b^2 = \frac{5}{9} a^2 \quad \text{(2)} \]

### Step 5: Substitute (2) into (1)
Now we substitute \( b^2 \) from equation (2) into equation (1):
\[ 2\left(\frac{5}{9} a^2\right) = 5a \]
This simplifies to:
\[ \frac{10}{9} a^2 = 5a \]
Multiplying both sides by 9 to eliminate the fraction:
\[ 10a^2 = 45a \]
Rearranging gives:
\[ 10a^2 - 45a = 0 \]
Factoring out \( 5a \):
\[ 5a(2a - 9) = 0 \]
Thus, \( a = 0 \) or \( a = \frac{9}{2} \). Since \( a \) cannot be zero, we have:
\[ a = \frac{9}{2} \]

### Step 6: Find \( b^2 \)
Now substituting \( a \) back into equation (2) to find \( b^2 \):
\[ b^2 = \frac{5}{9} \left(\frac{9}{2}\right)^2 = \frac{5}{9} \cdot \frac{81}{4} = \frac{45}{4} \]

### Step 7: Write the equation of the ellipse
The standard form of the equation of an ellipse centered at the origin is:
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]
Substituting \( a^2 = \left(\frac{9}{2}\right)^2 = \frac{81}{4} \) and \( b^2 = \frac{45}{4} \):
\[ \frac{x^2}{\frac{81}{4}} + \frac{y^2}{\frac{45}{4}} = 1 \]
Multiplying through by 4 to eliminate the denominators:
\[ \frac{4x^2}{81} + \frac{4y^2}{45} = 1 \]

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