Problem 200 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

If a hyperbola has length of its conjugate axis equal to 5 and the distance between its foci is 13 , then the eccentricity of the hyperbola is: (a) \(\frac{13}{12}\) (b) 2 (c) \(\frac{13}{6}\) (d) \(\frac{13}{8}\)

Step-by-Step Solution

Answer

The eccentricity of the hyperbola is \(\frac{13}{12}\), option (a).

Step 1: Understand Hyperbola Basics

A hyperbola consists of two fixed points known as the foci. The distance between the foci is \(2c\), where \(c\) is the distance from the center of the hyperbola to each focus. The eccentricity \(e\) is given by \(e = \frac{c}{a}\), where \(a\) is the semi-major axis. The conjugate axis has a length of \(2b\).

Step 2: Identify Given Values

In the problem, the length of the conjugate axis is given as 5, which means \(2b = 5\) or \(b = \frac{5}{2}\). The distance between the foci is 13, so \(2c = 13\) or \(c = \frac{13}{2}\).

Step 3: Apply the Hyperbola Relationship

For a hyperbola, the relationship between the axes dimensions is \(c^2 = a^2 + b^2\). We can substitute the known values of \(b\) and solve for \(a^2\).

Step 4: Calculate \(a^2\)

From the relation \(c^2 = a^2 + b^2\) and knowing \(c = \frac{13}{2}\) and \(b = \frac{5}{2}\), we plug in these values: \[\left(\frac{13}{2}\right)^2 = a^2 + \left(\frac{5}{2}\right)^2\]This simplifies to: \[\frac{169}{4} = a^2 + \frac{25}{4}\]Subtract \(\frac{25}{4}\) from both sides:\[a^2 = \frac{169}{4} - \frac{25}{4} = \frac{144}{4} = 36\]

Step 5: Calculate Eccentricity

Using the formula for eccentricity \(e = \frac{c}{a}\), substitute \(c = \frac{13}{2}\) and \(a = \sqrt{36} = 6\):\[e = \frac{\frac{13}{2}}{6} = \frac{13}{12}\]

Step 6: Confirm the Solution

Check the calculations:- Calculation of \(a^2\) is correct as each step logically follows from the given equations.- Eccentricity \(e = \frac{13}{12}\) fits within the choices, confirming our solution is correct.

Key Concepts

Eccentricity

Eccentricity is a key concept when studying the geometry of hyperbolas. It measures how "stretched out" or "flattened" the shape is. For a hyperbola, this is always greater than 1. The eccentricity, denoted as \( e \), is determined using the formula:

• \( e = \frac{c}{a} \)

Here, \( c \) represents the distance from the center to a focus, and \( a \) is the distance from the center to a vertex on the major axis. Hence, in hyperbolas, eccentricity helps in differentiating them from ellipses, for which \( e < 1 \).
Understanding eccentricity is crucial because it defines the hyperbola’s shape. The larger \( e \), the more 'spread out' it appears. In our example, through calculations, the eccentricity turned out to be \( \frac{13}{12} \). This value indicates that the hyperbola is relatively stretched, given it's barely more than 1.
In practical terms, this means the curve is symmetric about its center but extends vastly in opposite directions.

Conjugate Axis

The conjugate axis is an essential characteristic of a hyperbola that complements its primary direction of extension. While the major axis, which runs through the foci, is the longest dimension, the conjugate axis lies perpendicular to it.

• The length of the conjugate axis is given as \(2b\).

In our exercise, this length is mentioned to be 5, meaning \( b = \frac{5}{2} \). The importance of the conjugate axis lies in its relationship with the major axis. Together, they form a box-like structure that encloses the hyperbola, showcasing its full symmetric shape.
This axis, typically shorter than the major axis, does not cross the foci but crosses the hyperbola’s center. Although it doesn't serve as the primary spread of the hyperbola, it essentially determines the wideness of the box that circumscribes the two branches of the curve.
Understanding the conjugate axis is vital for calculating other characteristics like the eccentricity, using the equation \( c^2 = a^2 + b^2 \). This relationship ties together the linear dimensions of the hyperbola and reflects its geometric balance.

Distance Between Foci

The distance between the foci is a definitive characteristic of hyperbolas that sets them apart from other conic sections. This distance helps ascertain the space between the "two components" or branches of the hyperbola.

• It is expressed as \(2c\), where \(c\) is the distance from the center to one focus.

In the problem we examined, this distance was given as 13, leading to \( c = \frac{13}{2} \). This clear measure is crucial because it affects the hyperbola's shape and its properties, such as eccentricity.
In a hyperbola, a higher distance between foci typically points towards a more 'open' shape. However, it should be viewed relative to the semi-major axis \( a \).
Upon calculating \( a^2 \) and utilizing the formula \( c^2 = a^2 + b^2 \), we confirmed the values align with other attributes like conjugate axis and eccentricity. Understanding this relationship not only reinforces comprehension of the hyperbola's geometry but also aids in solving similar mathematical problems.