Question

If the eccentricity and the length of latus rectum of an ellipse are, respectively, \(\frac{1}{5}\) and \(\frac{48}{5}\), then the length of the major axis of the ellipse is

• A. 5
• B. 6
• C. 8
• D. 10
• E. 12

Hint:

Always start by expressing \(b^2\) in terms of \(a\) using the latus rectum formula. This usually leads to a direct linear equation for \(a\) when substituted into the eccentricity formula.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
We use the standard relationships for an ellipse to link eccentricity \(e\), latus rectum length, and semi-major axis \(a\).

Step 2: Key Formula or Approach:
1. Length of latus rectum = \(\frac{2b^2}{a}\).
2. Relation between \(a, b, e\): \(b^2 = a^2(1 - e^2)\).
3. Length of major axis = \(2a\).

Step 3: Detailed Explanation:
Given \(e = \frac{1}{5}\) and \(\frac{2b^2}{a} = \frac{48}{5}\).
From latus rectum:
\[ \frac{b^2}{a} = \frac{24}{5} \implies b^2 = \frac{24a}{5} \dots (i) \]
Substitute \(e\) into the relationship \(b^2 = a^2(1 - e^2)\):
\[ b^2 = a^2(1 - \frac{1}{25}) = a^2 \cdot \frac{24}{25} \dots (ii) \]
Equating (i) and (ii):
\[ \frac{24a}{5} = \frac{24a^2}{25} \]
Divide by \(24a\) (since \(a \neq 0\)):
\[ \frac{1}{5} = \frac{a}{25} \implies a = 5 \]
The length of the major axis is \(2a = 2 \cdot 5 = 10\).

Step 4: Final Answer:
The length of the major axis is 10.