Question

Let the eccentricity $e$ of a hyperbola satisfy the equation $6e^2 - 11e + 3 = 0$. If the foci of the hyperbola are $(3, 5)$ and $(3, -4)$, then the length of its latus rectum is :

• A. $11/3$
• B. $17/3$
• C. $15/2$
• D. $17/2$

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
We first solve for eccentricity $e$, use the distance between foci to find the transverse axis length $2a$, and then find $b$ to calculate the latus rectum $2b^2/a$.
: Key Formula or Approach:
Distance between foci $= 2ae$. Length of latus rectum $= 2b^2/a$. $b^2 = a^2(e^2 - 1)$.
Step 2: Detailed Explanation:
$6e^2 - 11e + 3 = 0 \implies (2e - 3)(3e - 1) = 0$.
Since for a hyperbola $e>1$, we take $e = 3/2$.
Foci are $(3, 5)$ and $(3, -4)$. Distance between them $= \sqrt{(3-3)^2 + (5 - (-4))^2} = 9$.
$2ae = 9 \implies 2a(3/2) = 9 \implies 3a = 9 \implies a = 3$.
Now, $b^2 = a^2(e^2 - 1) = 3^2((3/2)^2 - 1) = 9(9/4 - 1) = 9(5/4) = 45/4$.
Length of latus rectum $= \frac{2b^2}{a} = \frac{2(45/4)}{3} = \frac{45/2}{3} = \frac{15}{2}$.
Step 3: Final Answer:
The length of the latus rectum is $15/2$.