If distance between the foci of an ellipse is 6 and distance between its directrices is 12, then length of its latus rectum is : (A)4 (B)`3sqrt2` (C)9 (D)`2sqrt2`

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the given information
We know that:
- The distance between the foci of the ellipse is 6.
- The distance between the directrices is 12.

### Step 2: Relate the distance between the foci to the semi-major axis
The distance between the foci of an ellipse is given by \(2c\), where \(c = \sqrt{a^2 - b^2}\) and \(a\) is the semi-major axis. Therefore, we have:
\[ 2c = 6 \implies c = 3 \]

### Step 3: Relate the distance between the directrices to the semi-major axis and eccentricity
The distance between the directrices is given by \( \frac{2a}{e} \), where \(e\) is the eccentricity of the ellipse. We know that:
\[ \frac{2a}{e} = 12 \implies \frac{a}{e} = 6 \]

### Step 4: Express eccentricity in terms of \(a\)
From the relationship \(c = ae\), we can express \(e\) as:
\[ e = \frac{c}{a} \]
Substituting \(c = 3\):
\[ e = \frac{3}{a} \]

### Step 5: Substitute \(e\) in the equation from Step 3
Substituting \(e\) in the equation \( \frac{a}{e} = 6 \):
\[ \frac{a}{\frac{3}{a}} = 6 \implies \frac{a^2}{3} = 6 \implies a^2 = 18 \implies a = 3\sqrt{2} \]

### Step 6: Calculate \(b^2\)
Using the relationship \(c^2 = a^2 - b^2\):
\[ c^2 = 3^2 = 9 \quad \text{and} \quad a^2 = 18 \]
Thus,
\[ 9 = 18 - b^2 \implies b^2 = 9 \implies b = 3 \]

### Step 7: Calculate the length of the latus rectum
The length of the latus rectum \(L\) is given by the formula:
\[ L = \frac{2b^2}{a} \]
Substituting the values of \(b\) and \(a\):
\[ L = \frac{2 \cdot 9}{3\sqrt{2}} = \frac{18}{3\sqrt{2}} = \frac{6}{\sqrt{2}} = 3\sqrt{2} \]

### Final Answer
Thus, the length of the latus rectum is:
\[ \boxed{3\sqrt{2}} \]