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 \(\_\_\_\_\_\_\_\_\_\).
Correct Answer: 182
Approach Solution - 1
The problem asks for the value of \( l^2 + m^2 + n^2 \) based on the properties of a hyperbola whose latus rectum subtends a specific angle at the center.
Concept Used:
1. Standard Hyperbola: The equation of a horizontal hyperbola centered at the origin is \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \).
2. Latus Rectum: The latus rectum of a hyperbola is a chord passing through a focus and perpendicular to the transverse axis. - The foci are located at \( (\pm c, 0) \), where \( c^2 = a^2 + b^2 \). - The equations of the latus recta are \( x = \pm c \). - The endpoints of the latus rectum corresponding to the focus at \( (c, 0) \) are \( L(c, \frac{b^2}{a}) \) and \( L'(c, -\frac{b^2}{a}) \).
3. Angle Subtended at the Center: The angle subtended by the latus rectum at the center \( O(0,0) \) is the angle \( \angle LOL' \). Using trigonometry in the right-angled triangle formed by the center, a focus, and an endpoint of the latus rectum, we can relate the parameters of the hyperbola.
Step-by-Step Solution:
Step 1: Identify the parameters of the given hyperbola.
The equation of the hyperbola is \( \frac{x^2}{9} - \frac{y^2}{b^2} = 1 \). Comparing this with the standard form, we have \( a^2 = 9 \), so \( a = 3 \).
The relationship between \(a, b,\) and \(c\) is \( c^2 = a^2 + b^2 = 9 + b^2 \), which means \( c = \sqrt{9 + b^2} \).
The endpoints of the latus rectum passing through the focus \( (c, 0) \) are \( L(c, \frac{b^2}{a}) \) and \( L'(c, -\frac{b^2}{a}) \). Substituting \( a = 3 \), the points are \( L(c, \frac{b^2}{3}) \) and \( L'(c, -\frac{b^2}{3}) \).
Step 2: Use the angle condition to form an equation.
The latus rectum \( LL' \) subtends an angle of \( \frac{\pi}{3} \) at the center \( O(0,0) \). The x-axis is the axis of symmetry and bisects the angle \( \angle LOL' \). Let \( M \) be the focus on the x-axis, \( M(c, 0) \). In the right-angled triangle \( \triangle OML \), the angle \( \angle LOM \) is half of the total angle.
\[ \angle LOM = \frac{1}{2} \angle LOL' = \frac{1}{2} \times \frac{\pi}{3} = \frac{\pi}{6} \]In \( \triangle OML \), the slope of the line segment OL is given by \( \tan(\angle LOM) \).
\[ \tan(\angle LOM) = \frac{\text{length of opposite side (ML)}}{\text{length of adjacent side (OM)}} = \frac{b^2/a}{c} = \frac{b^2}{ac} \]Step 3: Solve the equation for \( b^2 \).
We have \( \tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}} \). Substituting the values of \(a\) and \(c\):
\[ \frac{1}{\sqrt{3}} = \frac{b^2}{3\sqrt{9 + b^2}} \]Rearranging the terms:
\[ 3\sqrt{9 + b^2} = \sqrt{3} b^2 \]Squaring both sides of the equation:
\[ (3\sqrt{9 + b^2})^2 = (\sqrt{3} b^2)^2 \] \[ 9(9 + b^2) = 3(b^2)^2 \]Divide by 3:
\[ 3(9 + b^2) = (b^2)^2 \] \[ 27 + 3b^2 = (b^2)^2 \]Let \( Y = b^2 \). The equation becomes a quadratic equation in \(Y\):
\[ Y^2 - 3Y - 27 = 0 \]Using the quadratic formula to solve for \(Y\):
\[ Y = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(1)(-27)}}{2(1)} = \frac{3 \pm \sqrt{9 + 108}}{2} = \frac{3 \pm \sqrt{117}}{2} \]Since \( Y = b^2 \) must be positive, we take the positive root:
\[ b^2 = \frac{3 + \sqrt{117}}{2} \]Simplifying the square root: \( \sqrt{117} = \sqrt{9 \times 13} = 3\sqrt{13} \).
\[ b^2 = \frac{3 + 3\sqrt{13}}{2} \]Step 4: Match the result with the given form and find \( l, m, n \).
The problem states that \( b^2 = \frac{l}{m} \left( 1 + \sqrt{n} \right) \). We can factor our result for \(b^2\):
\[ b^2 = \frac{3(1 + \sqrt{13})}{2} = \frac{3}{2}(1 + \sqrt{13}) \]Comparing this with the given form, we identify:
- \( l = 3 \)
- \( m = 2 \)
- \( n = 13 \)
We verify that \( l=3 \) and \( m=2 \) are coprime, as required.
Final Computation & Result:
Step 5: Calculate the value of \( l^2 + m^2 + n^2 \).
\[ l^2 + m^2 + n^2 = 3^2 + 2^2 + 13^2 \] \[ = 9 + 4 + 169 \] \[ = 13 + 169 = 182 \]The value of \( l^2 + m^2 + n^2 \) is 182.
Approach Solution -2
Given the hyperbola \( \frac{x^2}{9} - \frac{y^2}{b^2} = 1 \) with latus rectum subtending \( 60^\circ \) at the center, we have: \[ \tan 30^\circ = \frac{b^2 / a}{ae} = \frac{b^2}{a^2 e} = \frac{1}{\sqrt{3}} \]
This gives \( e = \frac{\sqrt{5}}{3} \). Using \( e^2 = 1 + \frac{b^2}{a^2} \):
\[ b^2 = 3b^4 + 27 \Rightarrow b^4 - 3b^2 - 27 = 0 \]
Solving, we get \( b^2 = \frac{1}{3}(1 + \sqrt{13}) \) with \( l = 2 \), \( m = 3 \), and \( n = 13 \).
Thus, \[ l^2 + m^2 + n^2 = 4 + 9 + 169 = 182 \]
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