If eccentricity of the ellipse `(x^(2))/(a^(2)+1)+(y^(2))/(a^(2)+2)=1` is `(1)/(sqrt6)`, then the ratio of the length of the latus rectum to the length of the major axis is

Options

A. `(5)/(6)`
B. `(3)/(sqrt6)`
C. `(2)/(3)`
D. `(2)/(sqrt6)`

Answer

To solve the problem, we need to find the ratio of the length of the latus rectum to the length of the major axis for the given ellipse. Let's break it down step by step.

### Step 1: Identify the equation of the ellipse
The given equation of the ellipse is:

\[ \frac{x^2}{a^2 + 1} + \frac{y^2}{a^2 + 2} = 1 \]

### Step 2: Define the semi-major and semi-minor axes
From the standard form of the ellipse, we can identify:

- \( b^2 = a^2 + 1 \) (denominator of \( x^2 \))
- \( a^2 = a^2 + 2 \) (denominator of \( y^2 \))

Since \( a^2 + 2 > a^2 + 1 \), we have \( a^2 + 2 \) as the semi-major axis and \( a^2 + 1 \) as the semi-minor axis.

### Step 3: Length of the latus rectum
The length of the latus rectum \( L \) for an ellipse is given by the formula:

\[ L = \frac{2b^2}{a} \]

### Step 4: Length of the major axis
The length of the major axis \( M \) is given by:

\[ M = 2a \]

### Step 5: Find the ratio of the latus rectum to the major axis
We need to find the ratio:

\[ \text{Ratio} = \frac{L}{M} = \frac{\frac{2b^2}{a}}{2a} = \frac{b^2}{a^2} \]

### Step 6: Use the eccentricity to find \( b^2/a^2 \)
We know the eccentricity \( e \) of the ellipse is given as:

\[ e = \frac{1}{\sqrt{6}} \]

The relationship between \( a^2 \), \( b^2 \), and \( e^2 \) is:

\[ b^2 = a^2(1 - e^2) \]

Substituting \( e^2 = \frac{1}{6} \):

\[ 1 - e^2 = 1 - \frac{1}{6} = \frac{5}{6} \]

Thus,

\[ b^2 = a^2 \cdot \frac{5}{6} \]

### Step 7: Substitute \( b^2 \) in the ratio
Now we substitute \( b^2 \) back into the ratio:

\[ \frac{b^2}{a^2} = \frac{a^2 \cdot \frac{5}{6}}{a^2} = \frac{5}{6} \]

### Conclusion
The ratio of the length of the latus rectum to the length of the major axis is:

\[ \frac{5}{6} \]