Problem 205 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

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 : (a) \(\frac{3}{2}\) (b) \(\sqrt{3}\) (c) 2 (d) \(\frac{2}{\sqrt{3}}\)

Step-by-Step Solution

Answer

The eccentricity of the hyperbola is \( \frac{2}{\sqrt{3}} \).

Step 1: Understanding the equation of the hyperbola

The general equation for a hyperbola with the center at the origin and transverse axis along the x-axis is \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \). Since the transverse axis length is 4, \( 2a = 4 \), so \( a = 2 \).

Step 2: Finding b^2 using given point

The hyperbola passes through the point \((4, 2)\). Substitute \(x = 4\) and \(y = 2\) into the hyperbola's equation to get \( \frac{4^2}{2^2} - \frac{2^2}{b^2} = 1 \). Simplifying gives \( \frac{16}{4} - \frac{4}{b^2} = 1 \) or \( 4 - \frac{4}{b^2} = 1 \).

Step 3: Solve for b^2

From \( 4 - \frac{4}{b^2} = 1 \), we get \( \frac{4}{b^2} = 3 \). Solving for \(b^2\), we have \( b^2 = \frac{4}{3} \).

Step 4: Finding the eccentricity

The eccentricity \(e\) of a hyperbola is given by \( e = \sqrt{1 + \frac{b^2}{a^2}} \). Substitute \(a^2 = 4\) and \(b^2 = \frac{4}{3}\) into this formula: \( e = \sqrt{1 + \frac{4/3}{4}} = \sqrt{1 + \frac{1}{3}} = \sqrt{\frac{4}{3}} \).

Step 5: Simplify the eccentricity

Simplify \( e = \sqrt{\frac{4}{3}} \) to \( e = \frac{2}{\sqrt{3}} \). Hence, the eccentricity of the hyperbola is \( \frac{2}{\sqrt{3}} \).

Key Concepts

Eccentricity

The eccentricity of a hyperbola is a critical measure that tells us how "stretched" it is. In simple terms, it's the factor that distinguishes a round shape (like a circle) from a more elongated one (a hyperbola, in this case). For circles, the eccentricity is a perfect 0, but hyperbolas have eccentricities greater than 1.

Eccentricity is calculated using the formula:

• \( e = \sqrt{1 + \frac{b^2}{a^2}} \) for hyperbolas.

In this situation, where we found \( a^2 = 4 \) and \( b^2 = \frac{4}{3} \), we substitute these values into the formula to get:
\( e = \sqrt{1 + \frac{4/3}{4}} = \sqrt{\frac{4}{3}}= \frac{2}{\sqrt{3}} \).

This tells us that the hyperbola is moderately stretched along its axis, giving it its characteristic shape. Understanding eccentricity can help you anticipate the behavior and appearance of hyperbolas even before sketching them.

Transverse Axis

The transverse axis is the segment that connects the two vertices of a hyperbola. It is crucial because it defines the orientation and basic dimensions of the hyperbola. Think of it as the main "spine" of the hyperbola along which it expands.

For hyperbolas centered at the origin, the transverse axis can either align with the x-axis or the y-axis. In the exercise provided, the transverse axis is along the x-axis and is given as having a length of 4.

Since the transverse axis is 4 units long, it implies that its half-length, denoted as \( a \), satisfies \( 2a = 4 \). Thus, \( a = 2 \).
This informs the setup of our hyperbola equation, ensuring it is correctly oriented and dimensioned according to its defined parameters.

Hyperbola Equation

The hyperbola equation helps us capture its geometric properties in a mathematical format. When a hyperbola is centered at the origin, with the transverse axis aligned with the x-axis, the standard form of its equation is:

• \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \).

In this case, the transverse axis being along the x-axis means that \( a^2 \) will directly influence the hyperbola's spread along the x-direction.

Given \( a = 2 \) — derived from the transverse axis length — we plug this into the equation, setting \( a^2 = 4 \).

To find \( b^2 \), which helps define the spread along the y-axis, we used the given point \((4, 2)\) on the hyperbola. Placing \((4, 2)\) in the equation gave us the necessary calculations to find \( b^2 = \frac{4}{3} \).

These elements come together to form the complete picture of the hyperbola, giving an ideal setup to interpret and solve problems related to its geometry.