The length of the latus rectum of the ellipse `2x^(2) + 3y^(2) - 4x - 6y - 13 = 0` is

Options

A. 5
B. 4
C. 8
D. 12

Answer

To find the length of the latus rectum of the ellipse given by the equation \(2x^2 + 3y^2 - 4x - 6y - 13 = 0\), we will follow these steps:

### Step 1: Rewrite the equation of the ellipse
We start with the equation:
\[ 2x^2 + 3y^2 - 4x - 6y - 13 = 0 \]
We can rearrange this to isolate the constant on one side:
\[ 2x^2 + 3y^2 - 4x - 6y = 13 \]

### Step 2: Complete the square for \(x\) and \(y\)
Next, we will complete the square for the \(x\) and \(y\) terms.

**For \(x\):**
\[ 2(x^2 - 2x) = 2((x - 1)^2 - 1) = 2(x - 1)^2 - 2 \]

**For \(y\):**
\[ 3(y^2 - 2y) = 3((y - 1)^2 - 1) = 3(y - 1)^2 - 3 \]

Now substituting these back into the equation gives:
\[ 2((x - 1)^2 - 1) + 3((y - 1)^2 - 1) = 13 \]
\[ 2(x - 1)^2 - 2 + 3(y - 1)^2 - 3 = 13 \]
\[ 2(x - 1)^2 + 3(y - 1)^2 - 5 = 13 \]
\[ 2(x - 1)^2 + 3(y - 1)^2 = 18 \]

### Step 3: Divide by 18 to get the standard form
Now we divide the entire equation by 18:
\[ \frac{2(x - 1)^2}{18} + \frac{3(y - 1)^2}{18} = 1 \]
This simplifies to:
\[ \frac{(x - 1)^2}{9} + \frac{(y - 1)^2}{6} = 1 \]

### Step 4: Identify \(a^2\) and \(b^2\)
From the standard form of the ellipse \(\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1\), we identify:
- \(a^2 = 9\) (thus \(a = 3\))
- \(b^2 = 6\) (thus \(b = \sqrt{6}\))

### Step 5: Calculate the length of the latus rectum
The formula for the length of the latus rectum \(L\) of an ellipse is given by:
\[ L = \frac{2b^2}{a} \]
Substituting the values we found:
\[ L = \frac{2 \cdot 6}{3} = \frac{12}{3} = 4 \]

### Conclusion
The length of the latus rectum of the ellipse is \(4\).

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