The lenth of the latus rectum of the ellipse `3x^2+y^2=12` is :

Options

A. `4`
B. `3`
C. `8`
D. `(4)/(sqrt(3))`

Answer

To find the length of the latus rectum of the ellipse given by the equation \(3x^2 + y^2 = 12\), we can follow these steps:

### Step 1: Rewrite the equation in standard form
We start with the equation of the ellipse:
\[ 3x^2 + y^2 = 12 \]
To convert this into standard form, we divide the entire equation by 12:
\[ \frac{3x^2}{12} + \frac{y^2}{12} = 1 \]
This simplifies to:
\[ \frac{x^2}{4} + \frac{y^2}{12} = 1 \]

### Step 2: Identify \(a^2\) and \(b^2\)
From the standard form \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\), we can identify:
- \(a^2 = 4\) (thus \(a = 2\))
- \(b^2 = 12\) (thus \(b = 2\sqrt{3}\))

### Step 3: Determine the relationship between \(a\) and \(b\)
Since \(b^2 > a^2\) (because \(12 > 4\)), this indicates that the ellipse is vertical.

### Step 4: Use the formula for the length of the latus rectum
The length of the latus rectum \(L\) of an ellipse is given by the formula:
\[ L = \frac{2a^2}{b} \]
Substituting the values of \(a^2\) and \(b\):
\[ L = \frac{2 \times 4}{2\sqrt{3}} = \frac{8}{2\sqrt{3}} = \frac{4}{\sqrt{3}} \]

### Step 5: Final answer
Thus, the length of the latus rectum of the ellipse \(3x^2 + y^2 = 12\) is:
\[ \frac{4}{\sqrt{3}} \]

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