If the distance between the foci of an ellipse is 6 and the distance between its directrices is 12,

Q: Wait, why is the distance between the directrices $\frac{2a}{e}$? I thought it was just $\frac{a}{e}$?

That is a common point of confusion! The directrices are two lines, one on the left and one on the right of the center. Their equations are $x = \frac{a}{e}$ and $x = -\frac{a}{e}$. The distance between these two vertical lines is the difference between their x-coordinates: $\frac{a}{e} - (-\frac{a}{e}) = \frac{2a}{e}$. Always remember that the directrices come in a pair!

Q: Can I use $b^2 = a^2 - c^2$ instead of $b^2 = a^2(1 - e^2)$?

Absolutely! They are mathematically identical. Since $c = ae$, then $c^2 = a^2e^2$. Substituting this into $b^2 = a^2 - c^2$ gives $b^2 = a^2 - a^2e^2$, which is exactly $a^2(1 - e^2)$. Using $b^2 = a^2 - c^2$ is often faster if you already have $c$ and $a$.

Q: Why do we have to rationalize the denominator in the final step?

Rationalizing the denominator is standard mathematical practice in JEE exams. It makes the expression cleaner and easier to compare with the given options. While $\frac{6}{\sqrt{2}}$ is numerically correct, $3\sqrt{2}$ is the simplified form that you will almost always find in the answer key.

Question

If the distance between the foci of an ellipse is 6 and the distance between its directrices is 12, then the length of its latus rectum is

Options:

(A) $2\sqrt{3}$

(B) $\sqrt{3}$

(C) $\frac{3}{\sqrt{2}}$

(D) $3\sqrt{2}$

Sigma Percentile · Accepted Answer

### The Geometry of the Ellipse: A Journey into Conic Sections

Welcome, future engineer! Today, we are going to peel back the layers of a classic JEE Advanced problem.

When you look at an ellipse, don't just see an oval. See a beautiful, balanced dance of parameters: the semi-major axis $a$, the semi-minor axis $b$, and the eccentricity $e$. These three variables define the very soul of the curve.

Our goal is to find the length of the latus rectum, a chord that captures the essence of the ellipse's width at its most critical points—the foci.

### Phase 1: Decoding the Geometry

Imagine you are standing at the center of the ellipse. The problem gives us two vital pieces of information.

First, the distance between the foci is $6$. We know the foci are located at $(\pm ae, 0)$. Thus, the total distance is $2ae = 6$, which simplifies to $ae = 3$. This is our first anchor point.

Next, we look at the directrices. These are the lines that 'govern' the ellipse's shape. Their equations are $x = \pm \frac{a}{e}$.

The distance between these two lines is $\frac{2a}{e} = 12$. Simplifying this, we get $\frac{a}{e} = 6$. Now, we have a system of two equations: $ae = 3$ and $\frac{a}{e} = 6$.

### Phase 2: The Algebraic Symphony

Many students would rush to solve for $e$ here, perhaps by dividing the equations. But look closer. If we multiply these two equations, the eccentricity $e$ vanishes entirely!

$$ (ae) 	imes \left( \frac{a}{e} ight) = 3 	imes 6 $$

$$ a^2 = 18 $$

This is the elegance of mathematics. By choosing the right operation, we bypass the complexity of solving for $e$ and jump straight to $a^2$.

Taking the square root, we find $a = \sqrt{18} = 3\sqrt{2}$. We have successfully pinned down the semi-major axis!

### Phase 3: Finding the Minor Axis

Now, we need the semi-minor axis $b$. We use the fundamental relationship $b^2 = a^2(1 - e^2)$.

Expanding this, we get $b^2 = a^2 - a^2e^2$. We already know $a^2 = 18$.

What about $a^2e^2$? That is simply $(ae)^2$. Since $ae = 3$, then $a^2e^2 = 9$. Substituting these values, we get $b^2 = 18 - 9 = 9$. The geometry is falling into place perfectly.

### Phase 4: The Final Calculation

The latus rectum is the chord passing through the focus, perpendicular to the major axis. Its length is defined by the formula $L = \frac{2b^2}{a}$.

We have all the pieces: $b^2 = 9$ and $a = 3\sqrt{2}$.

$$ L = \frac{2(9)}{3\sqrt{2}} = \frac{18}{3\sqrt{2}} $$

Simplifying this, we get $\frac{6}{\sqrt{2}}$. To make this look like our options, we rationalize the denominator by multiplying the numerator and denominator by $\sqrt{2}$:

$$ \frac{6\sqrt{2}}{2} = 3\sqrt{2} $$

And there it is! The length of the latus rectum is **$3\sqrt{2}$**. You have navigated the geometry, mastered the algebra, and arrived at the solution.

Animated Solution for Mathematics - Conic Sections: If the distance between the foci of an ellipse is 6 and the distance between its directrices is 12, then the length of its latus rectum is

Select Answer:

Visualized Solution

Standard Ellipse Equation

Distance Between Foci

Distance Between Directrices

Solving for $a^{2}$

Finding $a$

Eccentricity Relation

Calculating $b^{2}$

Latus Rectum Formula

Substituting Values

Final Calculation

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

The Geometry of the Ellipse

Phase 1

Phase 2

Phase 3

Phase 4

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