Question

Let $H$ be the hyperbola, whose foci are $(1 \pm \sqrt{2}, 0)$ and eccentricity is $\sqrt{2}$. Then the length of its latus rectum is _____

• A. 3
• B. $\frac{5}{2}$
• C. 2
• D. $\frac{3}{2}$

Hint:

For a hyperbola, the length of the latus rectum can be calculated using the formula \( L.R. = \frac{2b^2}{a} \), where \( b \) is the semi-minor axis and \( a \) is the semi-major axis.

Answer 1

The Correct Option is C

2ae=∣(1+2​)−(1+2​)∣=22​ 
ae=2​ 
a=1 
⇒b=1 
∵e=2​ 
⇒ Hyperbola is rectangular 
⇒L⋅R=a2b2​=2
So, the correct option is (C) : 2

Answer 2

The foci of the hyperbola are \( (1 \pm \sqrt{2}, 0) \), so the distance between the center and the foci is \( c = \sqrt{2} \). We are given that the eccentricity \( e = \sqrt{2} \). 
Step 1: We know that for a hyperbola, the relationship between the eccentricity, the distance from the center to the foci, and the semi-major axis is given by: \[ e = \frac{c}{a}. \] Thus: \[ \sqrt{2} = \frac{\sqrt{2}}{a} \quad \Rightarrow \quad a = 1. \] Step 2: For a hyperbola, we also know that: \[ b^2 = c^2 - a^2. \] Substitute \( c = \sqrt{2} \) and \( a = 1 \): \[ b^2 = (\sqrt{2})^2 - (1)^2 = 2 - 1 = 1 \quad \Rightarrow \quad b = 1. \] Step 3: The length of the latus rectum \( L.R. \) for a hyperbola is given by: \[ L.R. = \frac{2b^2}{a}. \] Substitute \( b = 1 \) and \( a = 1 \): \[ L.R. = \frac{2(1)^2}{1} = 2. \] Thus, the length of the latus rectum is \( 2 \).