Whose latus rectum is 5 and whose eccentricit is `(2)/(3)`

Answer

To find the equation of the ellipse given that its latus rectum is 5 and its eccentricity is \( \frac{2}{3} \), we can follow these steps:

### Step 1: Understand the formulas
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.

The eccentricity \( e \) of an ellipse is given by the formula:
\[ e = \sqrt{1 - \frac{B^2}{A^2}} \]

### Step 2: Set up the equations
From the problem, we know:
- \( L = 5 \)
- \( e = \frac{2}{3} \)

Using the latus rectum formula:
\[ 5 = \frac{2B^2}{A} \]
From this, we can express \( B^2 \):
\[ B^2 = \frac{5A}{2} \]

### Step 3: Substitute \( B^2 \) in the eccentricity formula
Now, substitute \( B^2 \) into the eccentricity formula:
\[ \frac{2}{3} = \sqrt{1 - \frac{B^2}{A^2}} \]
Substituting \( B^2 \):
\[ \frac{2}{3} = \sqrt{1 - \frac{\frac{5A}{2}}{A^2}} \]
This simplifies to:
\[ \frac{2}{3} = \sqrt{1 - \frac{5}{2A}} \]

### Step 4: Square both sides
Squaring both sides gives:
\[ \left(\frac{2}{3}\right)^2 = 1 - \frac{5}{2A} \]
\[ \frac{4}{9} = 1 - \frac{5}{2A} \]

### Step 5: Rearranging the equation
Rearranging the equation:
\[ \frac{5}{2A} = 1 - \frac{4}{9} \]
Calculating the right side:
\[ 1 - \frac{4}{9} = \frac{9}{9} - \frac{4}{9} = \frac{5}{9} \]
Thus, we have:
\[ \frac{5}{2A} = \frac{5}{9} \]

### Step 6: Solve for \( A \)
Cross-multiplying gives:
\[ 5 \cdot 9 = 5 \cdot 2A \]
\[ 45 = 10A \]
\[ A = \frac{45}{10} = \frac{9}{2} \]

### Step 7: Calculate \( B^2 \)
Now, substitute \( A \) back to find \( B^2 \):
\[ B^2 = \frac{5A}{2} = \frac{5 \cdot \frac{9}{2}}{2} = \frac{45}{4} \]

### Step 8: Write the equation of the ellipse
The standard form of the ellipse is:
\[ \frac{x^2}{A^2} + \frac{y^2}{B^2} = 1 \]
Substituting \( A^2 \) and \( B^2 \):
\[ \frac{x^2}{\left(\frac{9}{2}\right)^2} + \frac{y^2}{\left(\frac{\sqrt{45}}{2}\right)^2} = 1 \]
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
The equation of the ellipse is:
\[ \frac{x^2}{81} + \frac{y^2}{45} = 1 \]