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

Q: Wait, why is the distance between directrices $\frac{2a}{e}$?

The directrices of a standard ellipse are the vertical lines $x = \frac{a}{e}$ and $x = -\frac{a}{e}$. The distance between these two parallel lines is simply the difference of their x-coordinates: $\frac{a}{e} - (-\frac{a}{e}) = \frac{2a}{e}$. It is a standard property you should keep in your toolkit!

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

That formula is for a hyperbola! For an ellipse, the eccentricity $e$ is always less than 1, so $1 - e^2$ is positive. Always remember: for an ellipse, it is $b^2 = a^2(1 - e^2)$, and for a hyperbola, it is $b^2 = a^2(e^2 - 1)$.

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

Rationalizing the denominator is a standard mathematical convention. It makes the expression cleaner and easier to compare with the given options. $\frac{6}{\sqrt{2}}$ is mathematically equivalent to $3\sqrt{2}$, which is the simplified form.

Question

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

Options:

(A) 4

(B) $3\sqrt{2}$

(C) 9

(D) $2\sqrt{2}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on the coordinate plane, looking at an ellipse. It is not just a squashed circle; it is a shape defined by a beautiful, rigid set of relationships.

We are given two pieces of information: the distance between the foci is $6$, and the distance between the directrices is $12$. Our goal is to find the length of the latus rectum.

Let us start by grounding ourselves in the standard equation of an ellipse:

$$ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 $$

Here, $a$ is the semi-major axis, $b$ is the semi-minor axis, and $e$ is the eccentricity. These three variables are the DNA of our ellipse.

### Decoding the Geometry

First, let us look at the foci. The foci are located at $(\pm ae, 0)$. The distance between them is the span from $-ae$ to $ae$, which is $2ae$.

We are told this distance is $6$. So, we have our first equation:

$$ 2ae = 6 \implies ae = 3 $$

Next, consider the directrices. These are the vertical lines $x = \frac{a}{e}$ and $x = -\frac{a}{e}$. The distance between them is $\frac{2a}{e}$.

We are given that this distance is $12$. Simplifying this, we get:

$$ \frac{2a}{e} = 12 \implies \frac{a}{e} = 6 $$

### The Algebraic Dance

We have two variables, $a$ and $e$, trapped in these equations. If we multiply the two equations, the $e$ terms will cancel out:

$$ (ae) 	imes \left( \frac{a}{e} ight) = 3 	imes 6 \implies a^2 = 18 $$

If we divide the first by the second, the $a$ terms will cancel out:

$$ \frac{ae}{a/e} = \frac{3}{6} \implies e^2 = \frac{1}{2} $$

We have successfully isolated $a^2$ and $e^2$. This is the power of algebraic manipulation—turning a complex geometric problem into a simple arithmetic one.

### The Bridge to the Latus Rectum

Now, we need the length of the latus rectum, which is defined by the formula:

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

To find $b^2$, we use the fundamental eccentricity relation for an ellipse: $b^2 = a^2(1 - e^2)$. Substituting our known values, $a^2 = 18$ and $e^2 = \frac{1}{2}$, we get:

$$ b^2 = 18 \left( 1 - \frac{1}{2} ight) = 18 	imes \frac{1}{2} = 9 $$

We know $b^2 = 9$ and $a = \sqrt{18} = 3\sqrt{2}$. Plugging these into our $LR$ formula:

$$ LR = \frac{2 	imes 9}{3\sqrt{2}} = \frac{18}{3\sqrt{2}} = \frac{6}{\sqrt{2}} $$

Finally, 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, danced through the algebra, and arrived at the solution with precision.

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

Select Answer:

Visualized Solution

Visualizing the Ellipse

Distance Between Foci

Simplifying for $a e$

Distance Between Directrices

Simplifying for $\frac{a}{e}$

Solving for $a^{2}$

Solving for $e^{2}$

Relationship Between $a, b,$ and $e$

Calculating $b^{2}$

The Latus Rectum Formula

Final Calculation Setup

Simplifying the Result

Rationalizing and Final Answer

The Sigma Insight: Foci, Directrices, and Eccentricity

Analyzing the Setup

Decoding the Geometry

The Algebraic Dance

The Bridge to the Latus Rectum

Similar Questions

Let H be the hyperbola, whose foci are $(1 \pm 2, 0)$ and eccentricity is $2$ . Then the length of its latus rectum is

Let an ellipse with centre $(1, 0)$ and latus rectum of length $\frac{1}{2}$ have its major axis along x-axis. If its minor axis subtends an angle $6 0^{\circ}$ at the foci, then the square of the sum of the lengths of its minor and major axes is equal to _______.

Let $f (x) = x^{2} + 9, g (x) = \frac{x}{x - 9}$ and $a = f \circ g (10), b = g \circ f (3)$ . If $e$ and $l$ denote the eccentricity and the length of the latus rectum of the ellipse $\frac{x ^{2}}{a} + \frac{y ^{2}}{b} = 1$ , then $8 e^{2} + l^{2}$ is equal to.

Let the product of the focal distances of the point $P (4, 23)$ on the hyperbola $H : \frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$ be 32. Let the length of the conjugate axis of $H$ be $p$ and the length of its latus rectum be $q$ . Then $p^{2} + q^{2}$ is equal to _______

Let one focus of the hyperbola $H : \frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$ be at $(10, 0)$ and the corresponding directrix be $x = \frac{9}{10}$ . If $e$ and $l$ respectively are the eccentricity and the length of the latus rectum of $H$ , then $9 (e^{2} + l)$ is equal to:

Let $0 < θ < \frac{π}{2}$ . If the eccentricity of the hyperbola $\frac{x ^{2}}{c o s ^{2} θ} - \frac{y ^{2}}{s i n ^{2} θ} = 1$ is greater than 2, then the length of its latus rectum lies in the interval :

The eccentricity of the hyperbola whose length of the latus rectum is equal to 8 and the length of its conjugate axis is equal to half of the distance between its foci, is :

A focus of an ellipse is at the origin. The directrix is the line $x = 4$ and the eccentricity is $\frac{1}{2}$ . Then the length of the semi-major axis is

If the length of the minor axis of ellipse is equal to half of the distance between the foci, then the eccentricity of the ellipse is :