Question

Let foci of a hyperbola be (3, 5) and (3, -4). If eccentricity ‘e’ of the hyperbola satisfies the equation \( 3e^2 - 11e + 6 = 0 \), then the length of the latus rectum of the hyperbola is:

• A. 20
• B. 24
• C. 18
• D. 26

Hint:

The latus rectum can also be written as \( 2a(e^2 - 1) \). Once you have \( a \) and \( e \), you can skip calculating \( b^2 \) separately to save time.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The distance between the foci of a hyperbola is given by \( 2ae \). The latus rectum is calculated using the formula \( \frac{2b^2}{a} \), where \( b^2 = a^2(e^2 - 1) \). First, we solve the quadratic equation for the eccentricity \( e \).
Step 2: Key Formula or Approach:
1. Solve \( 3e^2 - 11e + 6 = 0 \).
2. Distance between foci \( S_1(3, 5) \) and \( S_2(3, -4) \) is \( 2ae \).
3. Use \( b^2 = a^2(e^2 - 1) \) and \( LR = \frac{2b^2}{a} \).
Step 3: Detailed Explanation:
1. Solve for \( e \): \( (3e - 2)(e - 3) = 0 \). Since \( e > 1 \) for a hyperbola, \( e = 3 \).
2. Distance between foci: \( |5 - (-4)| = 9 \). Thus, \( 2ae = 9 \).
3. With \( e = 3 \): \( 2a(3) = 9 \implies 6a = 9 \implies a = \frac{3}{2} \).
4. Find \( b^2 \): \( b^2 = \left(\frac{3}{2}\right)^2 (3^2 - 1) = \frac{9}{4} \times 8 = 18 \).
5. Length of Latus Rectum: \[ \frac{2 \times 18}{3/2} = \frac{36}{3/2} = 12 \times 2 = 24 \]
Step 4: Final Answer:
The length of the latus rectum is 24.