Problem 204 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

Let \(0<\theta<\frac{\pi}{2}\). If the eccentricity of the hyperbola \(\frac{x^{2}}{\cos ^{2} \theta}-\frac{y^{2}}{\sin ^{2} \theta}=1\) is greater than 2 , then the length of its latus rectum lies in the interval: (a) \((3, \infty)\) (b) \((3 / 2,2]\) (c) \((2,3]\) (d) \((1,3 / 2]\)

Step-by-Step Solution

Answer

The length of the latus rectum lies in the interval (3, ∞).

Step 1: Identify properties of the given hyperbola

The given hyperbola is \(\frac{x^2}{\cos^2 \theta} - \frac{y^2}{\sin^2 \theta} = 1\). This hyperbola is in standard form \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\), where \(a^2 = \cos^2 \theta\) and \(b^2 = \sin^2 \theta\).

Step 2: Determine the eccentricity condition

For a hyperbola, the eccentricity \(e\) is given by the formula \(e = \sqrt{1 + \frac{b^2}{a^2}}\). Substituting the values of \(a^2\) and \(b^2\), we have:\[e = \sqrt{1 + \frac{\sin^2 \theta}{\cos^2 \theta}} = \sqrt{1 + \tan^2 \theta} = \sec \theta\]The problem states that the eccentricity is greater than 2: \(\sec \theta > 2\).

Step 3: Solve the eccentricity condition inequality

Given \(\sec \theta = \frac{1}{\cos \theta} > 2\), we find:\[\cos \theta < \frac{1}{2}\]Since \(0 < \theta < \frac{\pi}{2}\), possible values for \(\theta\) where \(\cos \theta < \frac{1}{2}\) is in the range \(\frac{\pi}{3} < \theta < \frac{\pi}{2}\).

Step 4: Determine latus rectum length

The length of the latus rectum \(L\) is given by \(L = \frac{2b^2}{a}\). Since \(b^2 = \sin^2 \theta\) and \(a = \cos \theta\), we use:\[L = \frac{2\sin^2 \theta}{\cos \theta}\]Simplifying, we get \[L = 2\cdot \sin \theta \cdot \tan \theta\].

Step 5: Evaluate latus rectum interval

To find the range of \(L\), note that when \(\theta = \frac{\pi}{3}\), \(\sin \theta = \frac{\sqrt{3}}{2}\) and \(\tan \theta = \sqrt{3}\), so:\[L = 2 \cdot \frac{\sqrt{3}}{2} \cdot \sqrt{3} = 3\]As \(\theta\) approaches \(\frac{\pi}{2}\), \(\tan \theta\) approaches infinity, causing \(L\) to approach infinity. Thus, \(L\) is within the interval \((3, \infty)\).

Key Concepts

Eccentricity of Conic Sections

The concept of eccentricity is crucial in understanding the shape and nature of conic sections like ellipses, parabolas, and hyperbolas. Eccentricity, commonly denoted as \(e\), provides insight into how much a conic section deviates from being circular.

In the context of a hyperbola, eccentricity is calculated using the formula: \[e = \sqrt{1 + \frac{b^2}{a^2}}\] where \(a\) and \(b\) are the hyperbola's semi-major and semi-minor axes, respectively. For a given hyperbola \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\), the values of \(a^2\) and \(b^2\) can be determined to derive \(e\).

The hyperbola discussed in the exercise has the form \(\frac{x^2}{\cos^2 \theta} - \frac{y^2}{\sin^2 \theta} = 1\). Using the definition, the eccentricity is found to be \(\sec \theta\), which is greater than 2. This indicates that this particular hyperbola is more flattened than a standard hyperbola, signifying a deeper deviation from the circular path.

Latus Rectum of Hyperbola

The latus rectum of a hyperbola is an important feature that illustrates the width of the hyperbola at its focal points. Specifically, it is a line segment perpendicular to the transverse axis that passes through one of the foci. Understanding its length helps in visualizing the size of the hyperbola's opening.

For a hyperbola given by \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\), the latus rectum is given by the formula:\[L = \frac{2b^2}{a}\] In this problem, \(b^2 = \sin^2 \theta\) and \(a = \cos \theta\). Hence, the length \(L\) becomes:\[L = \frac{2\sin^2 \theta}{\cos \theta} = 2\cdot \sin \theta \cdot \tan \theta\]This relationship shows how the trigonometric ratios of the angle \(\theta\) influence the latus rectum's length.

From the calculations, for \(\theta = \frac{\pi}{3}\), \(L\) equals 3, and as \(\theta\) approaches \(\frac{\pi}{2}\), \(L\) tends towards infinity, placing \(L\) in the interval \((3, \infty)\). This finding confirms the increasing nature of \(L\) with larger values of \(\theta\).

Trigonometric Functions in Geometry

The use of trigonometric functions allows for a deeper understanding of geometric problems, particularly those involving conic sections like hyperbolas. In this exercise, trigonometric identities significantly simplify computations and reveal relationships between angles and lengths.

Key trigonometric functions include:

• Sine (\(\sin\theta\)), which relates to the opposite side and hypotenuse of a right-angled triangle.
• Cosine (\(\cos\theta\)), which denotes the adjacent side over hypotenuse.
• Tangent (\(\tan\theta\)), the ratio of sine over cosine.
• Secant (\(\sec\theta\)), representing the reciprocal of cosine.

These identities assist in transforming expressions involving \(\sin \), \(\cos \), \(\tan \), and \(\sec \) as seen in finding \(L = 2\cdot \sin \theta \cdot \tan \theta\). The hyperbolic identity \(\sec \theta = \frac{1}{\cos \theta}\) being greater than 2 set the stage for determining \(\theta\) in a specific range.

Using trigonometry in this manner allows for a deeper and more insightful way to solve geometric problems, providing a broader scope of solution strategies.