Question

If the latus rectum through one of the foci of a hyperbola \( \frac{x^2}{9} - \frac{y^2}{b^2} = 1 \) subtends a right angle at the farther vertex of the hyperbola, then \(b^2 = \)

• A. 4
• B. 16
• C. 25
• D. 27

Hint:

For conics, relations involving eccentricity 'e' often simplify calculations. The condition that the latus rectum subtends a right angle at the farther vertex of a hyperbola leads to the simple relation \(e=2\). From this, \(b^2=a^2(e^2-1)=a^2(3)\) can be found quickly.

Answer 1

The Correct Option is D

Step 1: Identify the key points of the hyperbola.
The equation is \( \frac{x^2}{9} - \frac{y^2}{b^2} = 1 \). So \(a^2=9 \implies a=3\).
The vertices are at \( V(a,0)=(3,0) \) and \( V'(-a,0)=(-3,0) \).
The foci are at \( (\pm c, 0) \), where \( c^2 = a^2+b^2 = 9+b^2 \). Let's consider the focus in the right half-plane, \(S(c,0)\).
The ends of the latus rectum through S are \( L(c, b^2/a) \) and \( L'(c, -b^2/a) \).
Step 2: Set up the condition for a right angle.
The latus rectum (the segment LL') subtends an angle at the farther vertex. The farther vertex from S(c,0) is \(V'(-3,0)\).
The angle \( \angle LV'L' \) is a right angle. This means the lines V'L and V'L' are perpendicular.
The product of their slopes must be -1.
Slope of V'L, \( m_1 = \frac{(b^2/a) - 0}{c - (-3)} = \frac{b^2/a}{c+3} \).
Slope of V'L', \( m_2 = \frac{(-b^2/a) - 0}{c - (-3)} = \frac{-b^2/a}{c+3} \).
Step 3: Solve the equation from the perpendicularity condition.
\( m_1 \cdot m_2 = -1 \).
\( \left( \frac{b^2/a}{c+3} \right) \left( \frac{-b^2/a}{c+3} \right) = -1 \).
\( \frac{-b^4/a^2}{(c+3)^2} = -1 \implies b^4 = a^2(c+3)^2 \).
Taking the square root: \( b^2 = a(c+3) \).
Substitute \(a=3\): \( b^2 = 3(c+3) \).
We also know \(c^2 = 9+b^2\). Substitute \(c = (b^2/3)-3\).
\((\frac{b^2}{3}-3)^2 = 9+b^2 \implies \frac{b^4}{9} - 2b^2 + 9 = 9+b^2 \implies \frac{b^4}{9} = 3b^2\).
Since \(b^2 \neq 0\), we can divide by \(b^2\), giving \(b^2/9 = 3 \implies b^2 = 27\).