The length of latus-rectum of the ellipse `3x^(2) + 4y^(2) - 6x + 8y - 5 = 0` is

Options

A. 3
B. `(3)/(2)`
C. `(sqrt(3))/(2)`
D. none of these

Answer

To find the length of the latus rectum of the ellipse given by the equation \( 3x^{2} + 4y^{2} - 6x + 8y - 5 = 0 \), we will follow these steps:

### Step 1: Rewrite the equation in standard form
We start with the equation:
\[ 3x^{2} + 4y^{2} - 6x + 8y - 5 = 0 \]
Rearranging gives:
\[ 3x^{2} - 6x + 4y^{2} + 8y = 5 \]

### Step 2: Complete the square for the \(x\) terms
For the \(x\) terms \(3x^{2} - 6x\):
1. Factor out the 3:
\[ 3(x^{2} - 2x) \]
2. Complete the square:
\[ x^{2} - 2x = (x - 1)^{2} - 1 \]
So,
\[ 3((x - 1)^{2} - 1) = 3(x - 1)^{2} - 3 \]

### Step 3: Complete the square for the \(y\) terms
For the \(y\) terms \(4y^{2} + 8y\):
1. Factor out the 4:
\[ 4(y^{2} + 2y) \]
2. Complete the square:
\[ y^{2} + 2y = (y + 1)^{2} - 1 \]
So,
\[ 4((y + 1)^{2} - 1) = 4(y + 1)^{2} - 4 \]

### Step 4: Substitute back into the equation
Now substitute the completed squares back into the equation:
\[ 3((x - 1)^{2} - 1) + 4((y + 1)^{2} - 1) = 5 \]
This simplifies to:
\[ 3(x - 1)^{2} - 3 + 4(y + 1)^{2} - 4 = 5 \]
Combining like terms gives:
\[ 3(x - 1)^{2} + 4(y + 1)^{2} = 12 \]

### Step 5: Divide by 12 to get the standard form
Dividing the entire equation by 12:
\[ \frac{3(x - 1)^{2}}{12} + \frac{4(y + 1)^{2}}{12} = 1 \]
This simplifies to:
\[ \frac{(x - 1)^{2}}{4} + \frac{(y + 1)^{2}}{3} = 1 \]

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

### Step 7: 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:
\[ L = \frac{2 \cdot 3}{2} = 3 \]

### Final Answer
The length of the latus rectum of the ellipse is \(3\).

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