A hyperbola has its centre at the origin, passes through the point (4,2) and has transverse axis of

Q: Wait, how do I know if the transverse axis is along the x-axis or y-axis?

Great question! The problem explicitly states the transverse axis is along the x-axis. This tells us the hyperbola is 'horizontal,' meaning the $x^2$ term is positive in the standard form $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$. If it were along the y-axis, the $y^2$ term would be positive.

Q: Why is the length of the transverse axis $2a$ and not just $a$?

Think of the hyperbola as two symmetric branches. The distance from the center $(0,0)$ to the vertex of one branch is $a$. Since there is a symmetric vertex on the other side at $(-a, 0)$, the total distance between the two vertices—the transverse axis—is $a + a = 2a$.

Q: Can $b^2$ ever be negative in this formula?

In the context of a standard hyperbola equation $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, $b^2$ represents a positive constant. If your calculation leads to a negative $b^2$, it usually means there is an arithmetic error in substituting the point or rearranging the terms. Stay calm and re-check your signs!

Q: Is there a shortcut for eccentricity?

While you can memorize $e = \sqrt{1 + \frac{b^2}{a^2}}$, the real 'shortcut' is understanding the geometry. Eccentricity measures how much the hyperbola deviates from a circle. For a hyperbola, $e$ must always be greater than 1. If you get a value less than 1, you know something went wrong!

Q: Why do we use the point $(4,2)$ at all?

The equation $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ defines a family of hyperbolas. By giving you a specific point $(4,2)$ that the curve must pass through, the problem 'pins' the hyperbola down, forcing a specific value for $b$ that satisfies the geometry.

Question

A hyperbola has its centre at the origin, passes through the point (4,2) and has transverse axis of length 4 along the x-axis. Then the eccentricity of the hyperbola is :

Options:

(A) $2/\sqrt{3}$

(B) $3/2$

(C) $\sqrt{3}$

(D) 2

Sigma Percentile · Accepted Answer

### The Geometry of Openness: Unveiling the Hyperbola

Imagine you are standing at the origin of a coordinate plane. You are tasked with drawing a curve that is perfectly symmetric, one that stretches infinitely toward the horizon, yet is anchored by a specific point in space. This is the essence of a hyperbola.

Today, we are going to dissect a problem that isn't just about plugging numbers into formulas; it is about understanding the structural integrity of a conic section.

### Phase 1: Defining the Skeleton

Every hyperbola has a 'skeleton'—its standard equation. We are told the center is at the origin $(0,0)$ and the transverse axis lies along the $x$-axis. This tells us immediately that our hyperbola is horizontal.

The standard form is given by:

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

Here, $a$ is the distance from the center to the vertex. The problem gives us the length of the transverse axis as $4$. Since the transverse axis is defined as $2a$, we have $2a = 4$, which simplifies beautifully to $a = 2$.

Now, our skeleton is starting to take shape: $a^2 = 4$. Substituting this back, we get:

$$\frac{x^2}{4} - \frac{y^2}{b^2} = 1$$

### Phase 2: The Constraint of the Point

Now, we introduce the point $(4,2)$. Think of this point as a 'nail' holding our hyperbola in place. For the hyperbola to pass through this point, the coordinates $(4,2)$ must satisfy our equation.

Let's substitute them in:

$$\frac{4^2}{4} - \frac{2^2}{b^2} = 1$$

This simplifies to $\frac{16}{4} - \frac{4}{b^2} = 1$, or $4 - \frac{4}{b^2} = 1$. I know that seeing a variable in the denominator can be intimidating, but take a deep breath. We are just solving for $b^2$.

Rearranging the terms, we get $4 - 1 = \frac{4}{b^2}$, which leads us to $3 = \frac{4}{b^2}$. Thus, $b^2 = \frac{4}{3}$. We have successfully pinned down the geometry of our curve!

### Phase 3: The Eccentricity—The Measure of 'Hyperbolicity'

Finally, we arrive at the eccentricity, $e$. The eccentricity is the soul of a conic section; it tells us how 'stretched' the curve is. For a hyperbola, the relationship between $a$, $b$, and $e$ is defined by the elegant formula:

$$e = \sqrt{1 + \frac{b^2}{a^2}}$$

We have our values: $a^2 = 4$ and $b^2 = \frac{4}{3}$. Let's plug them in with confidence:

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

Look at that fraction $\frac{\frac{4}{3}}{4}$. The $4$s cancel out perfectly, leaving us with $\frac{1}{3}$. The expression simplifies to $e = \sqrt{1 + \frac{1}{3}}$, which is $\sqrt{\frac{4}{3}}$.

Taking the square root, we find our final result:

**$$e = \frac{2}{\sqrt{3}}$$**

### Conclusion: The Beauty of the Result

We started with a vague description of a curve and ended with a precise, numerical value for its eccentricity. Notice how the math guided us? We didn't need to guess; we simply followed the geometric constraints.

Remember, in JEE Advanced, the math is not a hurdle—it is the language that describes the shape of the universe. Keep practicing, keep visualizing, and most importantly, keep falling in love with the process.

Animated Solution for Mathematics - Conic Sections: A hyperbola has its centre at the origin, passes through the point (4,2) and has transverse axis of length 4 along the x-axis. Then the eccentricity of the hyperbola is :

Select Answer:

Visualized Solution

Standard Equation of Hyperbola

Length of Transverse Axis

Updating the Equation

Point on the Hyperbola

Substituting Point $P (4, 2)$

Simplifying the Equation

Solving for $b^{2}$

Eccentricity Formula

Substituting $a^{2}$ and $b^{2}$

Simplifying the Fraction

Final Calculation of Eccentricity

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

The Geometry of Openness

Phase 1

Phase 2

Phase 3

Conclusion

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