What is the sum of the major and minor axes of the ellipse whose eccentricity is `(4)/(5)` and length of latus rectum is 14.4 units?

Options

A. 32 units
B. 48 units
C. 64 units
D. None of these

Answer

To find the sum of the major and minor axes of the ellipse given its eccentricity and length of the latus rectum, we can follow these steps:

### Step 1: Understand the parameters
- The eccentricity (e) of the ellipse is given as \( \frac{4}{5} \).
- The length of the latus rectum (L) is given as 14.4 units.

### Step 2: Use the formula for eccentricity
The formula for eccentricity of an ellipse is given by:
\[ e = \sqrt{1 - \frac{b^2}{a^2}} \]
where \( a \) is the semi-major axis and \( b \) is the semi-minor axis.

Substituting the value of eccentricity:
\[ \frac{4}{5} = \sqrt{1 - \frac{b^2}{a^2}} \]
Squaring both sides:
\[ \left(\frac{4}{5}\right)^2 = 1 - \frac{b^2}{a^2} \]
\[ \frac{16}{25} = 1 - \frac{b^2}{a^2} \]
Rearranging gives:
\[ \frac{b^2}{a^2} = 1 - \frac{16}{25} = \frac{9}{25} \]
Thus, we have:
\[ b^2 = \frac{9}{25} a^2 \]

### Step 3: Use the formula for the length of the latus rectum
The length of the latus rectum (L) of an ellipse is given by:
\[ L = \frac{2b^2}{a} \]
Substituting the value of L:
\[ 14.4 = \frac{2b^2}{a} \]
Substituting \( b^2 = \frac{9}{25} a^2 \):
\[ 14.4 = \frac{2 \cdot \frac{9}{25} a^2}{a} \]
This simplifies to:
\[ 14.4 = \frac{18}{25} a \]
Multiplying both sides by 25:
\[ 360 = 18a \]
Dividing by 18:
\[ a = 20 \]

### Step 4: Find the value of \( b \)
Using \( b^2 = \frac{9}{25} a^2 \):
\[ b^2 = \frac{9}{25} \cdot (20^2) = \frac{9}{25} \cdot 400 = 144 \]
Taking the square root:
\[ b = 12 \]

### Step 5: Calculate the lengths of the axes
The lengths of the major and minor axes are:
- Major axis \( = 2a = 2 \cdot 20 = 40 \)
- Minor axis \( = 2b = 2 \cdot 12 = 24 \)

### Step 6: Find the sum of the major and minor axes
Sum of the major and minor axes:
\[ 2a + 2b = 40 + 24 = 64 \]

### Final Answer
The sum of the major and minor axes of the ellipse is **64 units**.

---