Problem 212 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

Let a aand \(b\) respectively be the semitransverse and semiconjugate axes of a hyperbola whose eccentricity satisfies the equation \(9 \mathrm{e}^{2}-18 \mathrm{e}+5=0 .\) If \(\mathrm{S}(5,0)\) is a focus and \(5 \mathrm{x}=9\) is the corresponding directrix of this hyperbola, then \(a^{2}-b^{2}\) is equal to: \(\quad\) Online April 9, 2016] (a) \(-7\) (b) \(-5\) (c) \(\underline{5}\) (d) 7

Step-by-Step Solution

Answer

The correct answer is (a) \(-7\).

Step 1: Identify the Eccentricity Equation

We know that the eccentricity satisfies the equation: \(9e^2 - 18e + 5 = 0\). To find the eccentricity \(e\), we'll solve this quadratic equation.

Step 2: Solve for Eccentricity

Use the quadratic formula \(e = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a = 9\), \(b = -18\), and \(c = 5\). Calculate the discriminant: \((-18)^2 - 4 \cdot 9 \cdot 5 = 324 - 180 = 144\). This gives us two solutions for \(e\): \( e = \frac{18 \pm \sqrt{144}}{18}\). Simplifying, \( e = \frac{18 \pm 12}{18}\) gives us two potential solutions: \(e = \frac{30}{18} = \frac{5}{3}\) or \(e = \frac{6}{18} = \frac{1}{3}\). The location of the focus \(S(5,0)\) suggests \(e > 1\), so \(e = \frac{5}{3}\).

Step 3: Apply Hyperbola Properties

For a hyperbola centered at the origin with a horizontal transverse axis, the relationship \(e = \frac{c}{a}\) holds, where \(c\) is the distance from the center to the focus. Since \(S(5,0)\) is a focus, \(c = 5\). Using \(e = \frac{5}{3}\), we can find \(a\): \( \frac{5}{3} = \frac{5}{a} \Rightarrow a = 3\).

Step 4: Relate the Directrix to the Hyperbola

The equation of the directrix is given by \(5x = 9\), or \(x = \frac{9}{5}\). For a hyperbola, the distance from the center to the directrix \(d = \frac{a}{e}\). Using \(a = 3\) and \(e = \frac{5}{3}\), we have \(d = \frac{3}{\frac{5}{3}} = \frac{9}{5}\), matching the given condition.

Step 5: Compute \(a^2 - b^2\)

The standard form of a hyperbola is \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\). We know \(c^2 = a^2 + b^2\) for hyperbolas. We have \(c = 5\), \(a = 3\), so \(c^2 = 25\), \(a^2 = 9\). Therefore, \(b^2 = 25 - 9 = 16\). Thus, \(a^2 - b^2 = 9 - 16 = -7\).

Step 6: Final Step: Outcome

Based on the calculations, \(a^2 - b^2 = -7\). Thus, the correct answer to the problem is \(-7\), which corresponds to option (a).

Key Concepts

Eccentricity

Eccentricity is a fundamental concept used when dealing with conic sections like ellipses, parabolas, and hyperbolas. It is a measure of how much a conic section deviates from being circular. Eccentricity is denoted as \( e \), and it helps define the shape of the curve.

For hyperbolas, the eccentricity always satisfies \( e > 1 \). This indicates that a hyperbola is more elongated compared to other conic sections. In our problem, we use the eccentricity formula pertaining to hyperbolas, where \( e = \frac{c}{a} \). Here, \( c \) is the distance from the center to the focus, and \( a \) is the semitransverse axis length.

The quadratic equation for eccentricity here was given as \( 9e^2 - 18e + 5 = 0 \). Solving it, we obtained two potential solutions: \( e = \frac{5}{3} \) and \( e = \frac{1}{3} \). Since hyperbolas have \( e > 1 \), we determine \( e = \frac{5}{3} \) as the correct eccentricity for this problem. The concept of eccentricity not only helps determine the structure of the hyperbola but also verifies its parameters like the directrix.

Quadratic Equation

A quadratic equation is a second-degree polynomial equation in one variable and is usually written in the form \( ax^2 + bx + c = 0 \). In solving for properties of conic sections, these equations frequently arise.

In the context of our problem, the quadratic equation \( 9e^2 - 18e + 5 = 0 \) is used to find the eccentricity \( e \) of the hyperbola. To solve this, we apply the quadratic formula: \[e = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]where \( a = 9 \), \( b = -18 \), and \( c = 5 \). By calculating the discriminant \( b^2 - 4ac \), which equals \( 144 \), we confirm the equation has real roots. Subsequently, substituting in the quadratic formula yields the solutions \( e = \frac{5}{3} \) or \( e = \frac{1}{3} \).

It's crucial in mathematics to understand how to solve quadratic equations, as they appear in diverse areas, including conic sections, algebra, and calculus. This technical skill helps unravel complex properties or behaviors of geometric figures like the hyperbola.

Conic Sections

Conic sections are curves obtained by intersecting a double-napped cone with a plane. They are fundamental in geometry and consist of four main types: circles, ellipses, parabolas, and hyperbolas. Each type is characterized by its eccentricity, where

• Circle has \( e = 0 \).
• Ellipse has \( 0 < e < 1 \).
• Parabola has \( e = 1 \).
• Hyperbola has \( e > 1 \).

In this exercise, we focused on hyperbolas. Hyperbolas have two separate branches and possess properties like foci, directrices, and asymptotes.

For a hyperbola centered at the origin, the standard form is:\[\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\]The difference \( a^2 - b^2 \) often comes into play in calculations, determining various aspects such as how much the hyperbola deviates from a central shape.

Each conic section type has its own applications, from planetary orbits modeled by ellipses to hyperbolas showing up in navigation and physics. Understanding these curves' geometric and algebraic properties is essential in solving related math problems accurately.