Question

If the distance between the foci of an ellipse is 4 and its eccentricity is $\frac{1}{2}$, then the length of its latus rectum is

• A. 4
• B. 8
• C. 6
• D. 12
• E. 10

Hint:

Once you have $a$ and $e$, you can directly calculate the latus rectum as \( 2a(1 - e^2) \). Substituting $a=4$ and $e=1/2$ gives \( 2 \times 4 \times (3/4) = 6 \).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
We are given parameters of an ellipse: the distance between foci (\(2c = 2ae\)) and the eccentricity (\(e\)).
We need to find the length of the latus rectum, which is \(\frac{2b^2}{a}\).
Step 2: Key Formula or Approach:
Foci distance: \(2ae = 4\).
Use \(b^2 = a^2(1 - e^2)\) to find \(b^2\).
Latus rectum = \(\frac{2b^2}{a}\).
Step 3: Detailed Explanation:
Given \(2ae = 4\), so \(ae = 2\).
Since \(e = \frac{1}{2}\):
\[ a\left(\frac{1}{2}\right) = 2 \implies a = 4 \] Now, calculate \(b^2\):
\[ b^2 = a^2(1 - e^2) \] \[ b^2 = 4^2\left(1 - \left(\frac{1}{2}\right)^2\right) \] \[ b^2 = 16\left(1 - \frac{1}{4}\right) = 16\left(\frac{3}{4}\right) = 12 \] Now, calculate the length of the latus rectum:
\[ \text{Latus Rectum} = \frac{2b^2}{a} \] \[ = \frac{2(12)}{4} = \frac{24}{4} = 6 \] Step 4: Final Answer:
The length of the latus rectum is 6.