If the length of the latus rectum of an ellipse is 4 units and the distance between a focus and its

Q: Wait, why is the distance between the focus and the nearest vertex $a - ae$? I thought it was $ae - a$.

Great question! Remember that for an ellipse, the eccentricity $e$ is always less than $1$. Therefore, $ae$ is smaller than $a$. The vertex is at $(a, 0)$ and the focus is at $(ae, 0)$. Since $a > ae$, the distance is indeed $a - ae$. Always keep the geometry in mind to avoid sign errors!

Q: Why can't we just use the formula $e = \sqrt{1 - \frac{b^2}{a^2}}$ directly?

You absolutely can! If you solve for $a$ and $b$ first using the two given equations, you will get the same result. However, the method of substitution we used is often faster in competitive exams because it allows you to cancel out the variable $a$ without needing to find its specific value first. It is all about efficiency!

Q: Is it possible for $a$ to be zero? Why did we cancel it out?

In the context of an ellipse, $a$ represents the semi-major axis. If $a$ were zero, the ellipse would collapse into a single point, which contradicts the definition of an ellipse. Since $a$ must be a positive length, it is perfectly safe to divide both sides by $a$.

Question

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 $\frac{3}{2}$ units, then its eccentricity is :

Options:

(A) $\frac{1}{3}$

(B) $\frac{2}{3}$

(C) $\frac{1}{9}$

(D) $\frac{1}{2}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on the coordinate plane, looking at a perfect, symmetrical ellipse. It is not just a shape; it is a mathematical dance between two focal points and a major axis.

We are given two specific geometric clues: the length of the latus rectum is $4$ units, and the distance between a focus and its nearest vertex is $\frac{3}{2}$ units. Let us translate these geometric truths into the language of algebra.

### Phase 1: Translating the Clues

We consider a standard horizontal ellipse with the equation:

$$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, \quad 	ext{where } a > b$$

The vertex on the major axis is at $A(a, 0)$, and the focus is at $S(ae, 0)$. The length of the latus rectum is given by:

$$\frac{2b^2}{a} = 4$$

With a quick stroke of algebraic simplification, we find our first treasure:

$$b^2 = 2a$$

Next, we look at the distance between the focus $S(ae, 0)$ and the nearest vertex $A(a, 0)$. Geometrically, this is the difference in their x-coordinates: $a - ae$. We are told this distance is $\frac{3}{2}$, which gives us:

$$a(1 - e) = \frac{3}{2}$$

### Phase 2: The Fundamental Bridge

Now, we need to connect these pieces. The bridge that links the semi-major axis $a$, the semi-minor axis $b$, and the eccentricity $e$ is the fundamental identity of the ellipse:

$$b^2 = a^2(1 - e^2)$$

Let us substitute our first finding, $b^2 = 2a$, into this identity:

$$2a = a^2(1 - e^2)$$

### Phase 3: The Algebraic Dance

The term $(1 - e^2)$ is a classic difference of squares, which can be factored into $(1 - e)(1 + e)$. Our equation becomes:

$$2a = a^2(1 - e)(1 + e)$$

Now, observe that $a(1 - e)$ is present on the right side. Substituting our second clue, $a(1 - e) = \frac{3}{2}$, we get:

$$2a = a \cdot \left( \frac{3}{2} ight) \cdot (1 + e)$$

Since $a$ is the semi-major axis, it cannot be zero. We can confidently cancel $a$ from both sides to obtain:

$$2 = \frac{3}{2}(1 + e)$$

### The Final Revelation

The rest is a simple, satisfying conclusion. Multiply both sides by $\frac{2}{3}$ to isolate $(1 + e)$:

$$1 + e = \frac{4}{3}$$

Subtracting $1$ from both sides, we arrive at the final result:

**$$e = \frac{1}{3}$$**

The eccentricity of our ellipse is **$\frac{1}{3}$**. It is not just about finding the answer; it is about seeing how the variables interact, how they cancel, and how the geometry dictates the algebra.

Animated Solution for Mathematics - Conic Sections: 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 $\frac{3}{2}$ units, then its eccentricity is :

Select Answer:

Visualized Solution

Visualizing the Ellipse

The Latus Rectum

Latus Rectum Equation

Distance Between Focus and Vertex

Distance Equation

The Fundamental Relation

First Substitution

Strategic Factorization

Second Substitution

Canceling 'a'

Isolating 'e'

Final Calculation

Conclusion

The Sigma Insight: Standard Equations of Parabola, Ellipse, and Hyperbola

Analyzing the Setup

Phase 1

Phase 2

Phase 3

The Final Revelation

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