Ellipse
Latus Rectum and Eccentricity
Grade 11
Question:
<p>If the length of the latus rectum of an ellipse is 4 units and the distance between a focus and its nearest vertex on the major axis is \(\dfrac{3}{2}\) units, then its eccentricity is</p>
<p>\(\dfrac{1}{2}\)</p>
<p>\(\dfrac{1}{3}\)</p>
<p>\(\dfrac{2}{3}\)</p>
<p>\(\dfrac{1}{9}\)</p>
Step-by-Step Solution
Key Concept: Use the latus rectum formula L = 2b²/a and the distance condition (a - c = 3/2) together with c² = a² - b² to form a solvable system for eccentricity e = c/a.
<p><strong>Step 1:</strong> Use the latus rectum condition.</p><p>Length of latus rectum = 2b²/a = 4</p><p>Therefore: <strong>b² = 2a</strong> ... (i)</p><p><strong>Step 2:</strong> Use the distance condition.</p><p>Distance from focus to nearest major axis vertex = a - c = 3/2</p><p>Therefore: <strong>c = a - 3/2</strong> ... (ii)</p><p><strong>Step 3:</strong> Apply the ellipse relation c² = a² - b².</p><p>(a - 3/2)² = a² - 2a</p><p>a² - 3a + 9/4 = a² - 2a</p><p>-3a + 9/4 = -2a</p><p>-a = -9/4</p><p><strong>a = 9/4</strong></p><p><strong>Step 4:</strong> Find c and eccentricity.</p><p>c = a - 3/2 = 9/4 - 6/4 = 3/4</p><p>e = c/a = (3/4)/(9/4) = 3/9 = <strong>1/3</strong></p><p>∴ Answer: C</p>
Correct Answer: C