Question

Let \( H_n : \frac{x^2}{1 + n} + \frac{y^2}{3 + n} = 1, n \in \mathbb{N} \). Let \( k \) be the smallest even value of \( n \) such that the eccentricity of \( H_n \) is a rational number. If \( l \) is the length of the latus rectum of \( H_k \), then 21 \( l \) is equal to:

Hint:

For hyperbolas, the length of the latus rectum is \( \frac{2b^2}{a} \), where \( a \) and \( b \) are the semi-major and semi-minor axes, respectively.

Answer 1

Correct Answer: 306

The equation of the hyperbola is: \[ H_n: \frac{x^2}{1 + n} + \frac{y^2}{3 + n} = 1 \] To find the eccentricity \( e \) of this hyperbola, we use the formula: \[ e = \sqrt{ \frac{b^2}{a^2} + 1 } \] where \( a^2 = 1 + n \) and \( b^2 = 3 + n \). Therefore, \[ e = \sqrt{ \frac{3 + n}{1 + n} } \] We need \( e \) to be a rational number, so the ratio \( \frac{3 + n}{1 + n} \) should be a perfect square. To satisfy this, the smallest value of \( n \) such that \( e \) is rational is \( n = 48 \). Now, substituting \( n = 48 \) into the equations for \( a \) and \( b \): \[ a^2 = 1 + 48 = 49 \quad \text{and} \quad b^2 = 3 + 48 = 51 \] Thus, \[ a = 7 \quad \text{and} \quad b = \sqrt{51} \] The length of the latus rectum \( l \) of the hyperbola is given by: \[ l = \frac{2b^2}{a} \] Substituting the values for \( a \) and \( b \): \[ l = \frac{2 \times 51}{7} = \frac{102}{7} \] Finally, to find \( 21l \), we multiply \( l \) by 21: \[ 21l = 21 \times \frac{102}{7} = 306 \] Thus, the value of \( 21l \) is \( \boxed{306} \).