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 foci are given as \(S_1(3, 5)\) and \(S_2(3, -4)\). Since the \(x\)-coordinates are the same, the transverse axis is vertical.
Step 2: Key Formula or Approach:
Distance \(S_1S_2 = |5 - (-4)| = 9\).
So, \(2ae = 9\).
We are given the quadratic equation for eccentricity \(e\):
\[ 3e^2 - 11e + 6 = 0 \]
\[ 3e^2 - 9e - 2e + 6 = 0 \]
\[ 3e(e - 3) - 2(e - 3) = 0 \Rightarrow (3e - 2)(e - 3) = 0 \]
Since \(e>1\) for a hyperbola, we take \(e = 3\) (rejecting \(e = 2/3\)).
Step 3: Detailed Explanation:
Substitute \(e = 3\) into \(2ae = 9\):
\[ 2a(3) = 9 \Rightarrow 6a = 9 \Rightarrow a = \frac{3}{2} \]
For a hyperbola, \(b^2 = a^2(e^2 - 1)\).
\[ b^2 = \left(\frac{3}{2}\right)^2 (3^2 - 1) = \frac{9}{4} \times 8 = 18 \]
The length of the Latus Rectum (LR) is given by \(\frac{2b^2}{a}\):
\[ \text{LR} = \frac{2(18)}{3/2} = \frac{36 \times 2}{3} = 12 \times 2 = 24 \]
Step 4: Final Answer:
The length of the latus rectum is 24.