Question

Let the latus rectum of the hyperbola \( \frac{x^2}{9} - \frac{y^2}{b^2} = 1 \) subtend an angle of \( \frac{\pi}{3} \) at the center of the hyperbola. If \( b^2 \) is equal to \( \frac{1}{m} \left( 1 + \sqrt{n} \right) \), where \( l \) and \( m \) are coprime numbers, then \( l^2 + m^2 + n^2 \) is equal to \(\_\_\_\_\_\_\_\_\_\).

Answer 1

Correct Answer: 182

The objective is to determine the value of \( l^2 + m^2 + n^2 \) given properties of a hyperbola, specifically the angle subtended by its latus rectum at the center.

Key Concepts:

1. Standard Hyperbola Equation: A horizontal hyperbola centered at the origin is represented by \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \).

2. Latus Rectum Definition: The latus rectum is a chord through a focus, perpendicular to the transverse axis. - Foci are at \( (\pm c, 0) \), where \( c^2 = a^2 + b^2 \). - Latus recta equations are \( x = \pm c \). - For the focus \( (c, 0) \), latus rectum endpoints are \( (c, \frac{b^2}{a}) \) and \( (c, -\frac{b^2}{a}) \).

3. Angle at Center: The angle subtended by the latus rectum at the center \( O(0,0) \) can be calculated using trigonometry. A right-angled triangle formed by the center, a focus, and a latus rectum endpoint relates the hyperbola's parameters.

Solution Breakdown:

Step 1: Parameter Identification.

Given hyperbola: \( \frac{x^2}{9} - \frac{y^2}{b^2} = 1 \). From this, \( a^2 = 9 \), so \( a = 3 \).

The relationship \( c^2 = a^2 + b^2 \) becomes \( c^2 = 9 + b^2 \), thus \( c = \sqrt{9 + b^2} \).

Latus rectum endpoints for focus \( (c, 0) \) are \( (c, \frac{b^2}{3}) \) and \( (c, -\frac{b^2}{3}) \).

Step 2: Angle Condition Formulation.

The latus rectum subtends \( \frac{\pi}{3} \) at the center. The x-axis bisects this angle. In right triangle \( \triangle OML \) (where \( M \) is focus \( (c, 0) \) and \( L \) is \( (c, \frac{b^2}{a}) \)), \( \angle LOM = \frac{\pi}{6} \).

\[\tan(\angle LOM) = \frac{\text{ML}}{\text{OM}} = \frac{b^2/a}{c} = \frac{b^2}{ac}\]

Step 3: Solving for \( b^2 \).

Using \( \tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}} \) and substituting \( a=3 \) and \( c=\sqrt{9 + b^2} \):

\[\frac{1}{\sqrt{3}} = \frac{b^2}{3\sqrt{9 + b^2}}\]

This leads to \( 3\sqrt{9 + b^2} = \sqrt{3} b^2 \). Squaring both sides yields \( 9(9 + b^2) = 3b^4 \), which simplifies to \( 27 + 3b^2 = b^4 \).

Letting \( Y = b^2 \), we get the quadratic equation \( Y^2 - 3Y - 27 = 0 \).

Solving for \( Y \) using the quadratic formula: \( Y = \frac{3 \pm \sqrt{9 - 4(1)(-27)}}{2} = \frac{3 \pm \sqrt{117}}{2} \).

Since \( b^2 \) must be positive, \( b^2 = \frac{3 + \sqrt{117}}{2} = \frac{3 + 3\sqrt{13}}{2} \).

Step 4: Matching Form and Identifying \( l, m, n \).

The given form is \( b^2 = \frac{l}{m} \left( 1 + \sqrt{n} \right) \). Our result is \( b^2 = \frac{3}{2}(1 + \sqrt{13}) \).

By comparison:

• \( l = 3 \)
• \( m = 2 \)
• \( n = 13 \)

\( l=3 \) and \( m=2 \) are coprime.

Final Calculation:

Step 5: Computing \( l^2 + m^2 + n^2 \).

\[l^2 + m^2 + n^2 = 3^2 + 2^2 + 13^2 = 9 + 4 + 169 = 182\]

The computed value of \( l^2 + m^2 + n^2 \) is 182.