Let H be the hyperbola, whose foci are (1 ± √2, 0) and eccentricity is √2. Then the length of its la

Q: Why is the center of the hyperbola $(1, 0)$?

The center of a hyperbola is the midpoint of the line segment connecting its two foci. Given the foci $S_1(1 + \sqrt{2}, 0)$ and $S_2(1 - \sqrt{2}, 0)$, the midpoint is $(\frac{(1 + \sqrt{2}) + (1 - \sqrt{2})}{2}, \frac{0 + 0}{2}) = (1, 0)$. It is a beautiful example of how symmetry simplifies our work!

Q: Why is the distance between the foci $2ae$?

In the standard form of a hyperbola, the distance from the center to each focus is $ae$. Since the hyperbola is symmetric, the distance from the center to one focus is $ae$, and the distance to the other is also $ae$. Thus, the total distance between the two foci is $ae + ae = 2ae$.

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

Absolutely! In a hyperbola, $c$ represents the distance from the center to the focus, which is $ae$. So, $c = ae$. Substituting this into $c^2 = a^2 + b^2$, we get $(ae)^2 = a^2 + b^2$, which simplifies to $a^2e^2 = a^2 + b^2$, or $b^2 = a^2(e^2 - 1)$. Both paths lead to the same result!

Question

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

Options:

(A) 2

(B) 3

(C) $\frac{5}{2}$

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

Sigma Percentile · Accepted Answer

### The Elegant Geometry of the Hyperbola

Welcome, fellow traveler on the path to JEE mastery! Today, we are going to peel back the layers of a classic hyperbola problem.

A hyperbola is not just an equation; it is a locus of points defined by its foci and its eccentricity. Let's embark on this journey to find the length of its latus rectum.

### Phase 1: Unlocking the Center

We start with the most fundamental clues: the foci, $S_1(1 + \sqrt{2}, 0)$ and $S_2(1 - \sqrt{2}, 0)$. Notice something beautiful? The $y$-coordinates are both zero.

This tells us immediately that our hyperbola is oriented along the $x$-axis. The center of any hyperbola is the midpoint of the segment connecting its foci.

By calculating the average of the $x$-coordinates, we find that the $\sqrt{2}$ terms cancel out perfectly:

$$C = \left( \frac{1+\sqrt{2}+1-\sqrt{2}}{2}, 0 ight) = (1, 0)$$

We have found our center! It is the anchor point from which the entire curve expands.

### Phase 2: The Dance of $a$ and $e$

Now, let's look at the distance between these two foci. In the language of conic sections, the distance from the center to a focus is defined as $ae$.

Since there are two foci, the total distance between them is $2ae$. Looking at our coordinates, the distance is $(1 + \sqrt{2}) - (1 - \sqrt{2}) = 2\sqrt{2}$.

Setting these equal, we get $2ae = 2\sqrt{2}$, which simplifies to $ae = \sqrt{2}$. We are given the eccentricity $e = \sqrt{2}$.

Substituting this into our equation, we get $a(\sqrt{2}) = \sqrt{2}$. With a simple division, we find $a = 1$. The semi-transverse axis is unity.

### Phase 3: The Hyperbola's Identity

To find the latus rectum, we need the semi-conjugate axis, $b$. The relationship connecting $a$, $b$, and $e$ is the DNA of the hyperbola:

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

This formula is your best friend in the exam hall. Let's plug in our values:

$$b^2 = (1)^2((\sqrt{2})^2 - 1)$$

Calculating the square of $\sqrt{2}$ gives us $2$, and $2 - 1 = 1$. Thus, $b^2 = 1$. Everything is falling into place with such satisfying precision.

### Phase 4: The Final Calculation

The latus rectum is the chord passing through the focus, perpendicular to the transverse axis. Its length is given by the elegant formula:

$$	ext{Length} = \frac{2b^2}{a}$$

We have all the pieces of the puzzle: $b^2 = 1$ and $a = 1$. Substituting these in, we get:

$$	ext{Length} = \frac{2(1)}{1} = 2$$

The length of the latus rectum is exactly **2**.

Isn't it marvelous? We started with just two points and an eccentricity, and through the symmetry of the hyperbola, we arrived at a clean, integer result. Keep this systematic approach in your toolkit, and no conic section will ever intimidate you again.

Animated Solution for Mathematics - Conic Sections: Let H be the hyperbola, whose foci are $(1 \pm 2, 0)$ and eccentricity is $2$ . Then the length of its latus rectum is

Select Answer:

Visualized Solution

Identifying the Foci

Center of the Hyperbola

Distance Between Foci

Calculating $a e$

Utilizing Eccentricity

Finding the Semi-Transverse Axis $a$

Relation Between $a, b,$ and $e$

Substituting $a$ and $e$

Calculating $b^{2}$

Formula for Latus Rectum

Final Length of Latus Rectum

The Sigma Insight: Foci, Directrices, and Eccentricity

The Elegant Geometry of the Hyperbola

Phase 1

Phase 2

Phase 3

Phase 4

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Animated Solution for Mathematics - Conic Sections: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let H be the hyperbola, whose foci are (1±2​,0) and eccentricity is 2​. Then the length of its latus rectum is