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

Answer

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

### Step 1: Understand the properties of the 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 2: Use the given latus rectum to find \( a \)
We know the latus rectum is 5, so we can set up the equation:
\[ \frac{2b^2}{a} = 5 \]
This can be rearranged to:
\[ 2b^2 = 5a \]
or
\[ b^2 = \frac{5a}{2} \]

### Step 3: Use the given eccentricity to find a relationship between \( a \) and \( b \)
The eccentricity \( e \) of an ellipse is given by:
\[ e = \frac{c}{a} \]
where \( c = \sqrt{a^2 - b^2} \). Given that \( e = \frac{2}{3} \), we can write:
\[ \frac{c}{a} = \frac{2}{3} \]
This implies:
\[ c = \frac{2a}{3} \]

### Step 4: Relate \( c \) to \( a \) and \( b \)
From the relationship \( c = \sqrt{a^2 - b^2} \), we substitute for \( c \):
\[ \frac{2a}{3} = \sqrt{a^2 - b^2} \]
Squaring both sides gives:
\[ \left(\frac{2a}{3}\right)^2 = a^2 - b^2 \]
This simplifies to:
\[ \frac{4a^2}{9} = a^2 - b^2 \]
Rearranging gives:
\[ b^2 = a^2 - \frac{4a^2}{9} \]
\[ b^2 = \frac{9a^2 - 4a^2}{9} = \frac{5a^2}{9} \]

### Step 5: Set the two expressions for \( b^2 \) equal to each other
From Step 2, we have \( b^2 = \frac{5a}{2} \) and from Step 4, we have \( b^2 = \frac{5a^2}{9} \). Setting these equal gives:
\[ \frac{5a}{2} = \frac{5a^2}{9} \]
Dividing both sides by 5 (assuming \( a \neq 0 \)):
\[ \frac{a}{2} = \frac{a^2}{9} \]
Cross-multiplying gives:
\[ 9a = 2a^2 \]
Rearranging gives:
\[ 2a^2 - 9a = 0 \]
Factoring out \( a \):
\[ a(2a - 9) = 0 \]
Thus, \( a = 0 \) or \( a = \frac{9}{2} \). Since \( a \) cannot be zero, we have:
\[ a = \frac{9}{2} \]

### Step 6: Substitute \( a \) back to find \( b^2 \)
Using \( a = \frac{9}{2} \) in the equation \( b^2 = \frac{5a}{2} \):
\[ b^2 = \frac{5 \cdot \frac{9}{2}}{2} = \frac{45}{4} \]

### Step 7: Write the equation of the ellipse
The standard form of the ellipse centered at the origin with axes along the coordinate axes 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 gives:
\[ \frac{4x^2}{81} + \frac{4y^2}{45} = 1 \]

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