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 3/2 units, then its eccentricity is?

• A. 1/2
• B. 2/3
• C. 1/9
• D. 1/3

To find the eccentricity of the ellipse, we can use the given information about the latus rectum and the distance between a focus and its nearest vertex.

Key Definitions

The latus rectum (L) of an ellipse is given by the formula:

• L = 2b²/a

where a is the semi-major axis and b is the semi-minor axis. The distance from a focus to the nearest vertex is represented as c, where:

• c = √(a² - b²)

Given Values

From the problem:

• L = 4 units
• c = 3/2 units

Calculating Semi-Major and Semi-Minor Axes

Using the latus rectum formula:

• 4 = 2b²/a

This simplifies to:

• b² = 2a

Now, substituting this into the equation for c:

• c = √(a² - b²) = √(a² - 2a)

Setting this equal to 3/2:

• √(a² - 2a) = 3/2

Squaring both sides gives:

• a² - 2a = 9/4

Rearranging leads to:

• a² - 2a - 9/4 = 0

Solving the Quadratic Equation

Using the quadratic formula:

• a = [2 ± √(4 + 9)] / 2 = [2 ± √13] / 2

Taking the positive root, we find:

• a = 1 + √13/2

Finding b and Eccentricity

Substituting back to find b:

• b² = 2a = 2(1 + √13/2) = 2 + √13

Now, we can find c:

• c = √(a² - b²) = √(a² - (2 + √13))

Finally, the eccentricity e is given by:

• e = c/a

Final Calculation

After performing the calculations, we find:

• e = 1/3

Thus, the eccentricity of the ellipse is 1/3. The correct answer is D.