Let `H` be the hyperbola,whose foci are `(1+-sqrt(2),0)` and eccentricity is `sqrt(2)`.Then the length of its latus rectum is :

Options

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

Answer

To find the length of the latus rectum of the hyperbola \( H \) with given foci and eccentricity, we can follow these steps:

### Step 1: Identify the foci and eccentricity
The foci of the hyperbola are given as \( (1+\sqrt{2}, 0) \) and \( (1-\sqrt{2}, 0) \). The eccentricity \( e \) is given as \( \sqrt{2} \).

### Step 2: Calculate the distance between the foci
The distance between the foci \( 2c \) can be calculated as:
\[ 2c = |(1+\sqrt{2}) - (1-\sqrt{2})| = |2\sqrt{2}| \]
Thus, we have:
\[ c = \sqrt{2} \]

### Step 3: Relate eccentricity, \( a \), and \( c \)
For hyperbolas, the relationship between the semi-major axis \( a \), the distance to the foci \( c \), and eccentricity \( e \) is given by:
\[ e = \frac{c}{a} \]
Substituting the values we have:
\[ \sqrt{2} = \frac{\sqrt{2}}{a} \]
From this, we can solve for \( a \):
\[ a = 1 \]

### Step 4: Calculate \( b \) using the relationship \( c^2 = a^2 + b^2 \)
We know:
\[ c^2 = a^2 + b^2 \]
Substituting the known values:
\[ (\sqrt{2})^2 = 1^2 + b^2 \]
This simplifies to:
\[ 2 = 1 + b^2 \]
Thus:
\[ b^2 = 1 \quad \Rightarrow \quad b = 1 \]

### Step 5: Calculate the length of the latus rectum
The length of the latus rectum \( L \) for a hyperbola is given by:
\[ L = \frac{2b^2}{a} \]
Substituting the values of \( b \) and \( a \):
\[ L = \frac{2 \cdot 1^2}{1} = 2 \]

### Conclusion
The length of the latus rectum of the hyperbola \( H \) is \( 2 \).

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