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

To find the length of the latus rectum of a hyperbola, we need to understand its properties and given parameters. Here we are provided with a hyperbola \( H \) that has:

• Foci at \( \left(1 \pm \sqrt{2}, 0\right) \)
• Eccentricity \( e = \sqrt{2} \)

The standard form of a hyperbola with a horizontal transverse axis is represented by the equation:

\(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\)

For this hyperbola:

• The center \((h, k)\) is at \((1, 0)\) since the foci are at \(1 \pm \sqrt{2}\).
• The distance between the center and each focus is \(c = \sqrt{2}\).

The relationship between the semi-major axis \(a\), semi-minor axis \(b\), and foci distance \(c\) for a hyperbola is given by \(c^2 = a^2 + b^2\). Also, the eccentricity \(e\) is defined as \(e = \frac{c}{a}\).

Given \(e = \sqrt{2}\), we have:

\(\sqrt{2} = \frac{\sqrt{2}}{a} \Rightarrow a = 1\)

Using \(c^2 = a^2 + b^2\):

\((\sqrt{2})^2 = 1^2 + b^2 \Rightarrow 2 = 1 + b^2 \Rightarrow b^2 = 1\)\)

Now, the formula for the length of the latus rectum \(L\) of a hyperbola is given by:

\(L = \frac{2b^2}{a}\)

Substituting the values of \(b^2\) and \(a\):

\(L = \frac{2 \times 1}{1} = 2\)

Thus, the length of the latus rectum is 2.