Question: Let $0 < \theta < {\pi \over 2}$. If the eccentricity of the hyperbola ${{{x^2}} \over {{{\cos }^2}\theta }} - {{{y^2}} \over {{{\sin }^2}\theta }}$ = 1 is greater than 2, then the length of its latus rectum lies in the interval:

Options: (A) (3, $\infty$) (B) $\left( {{3 \over 2},2} \right]$ (C) $\left( {1,{3 \over 2}} \right]$ (D) $\left( {2,3} \right]$

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This problem requires a strong understanding of the standard form of a hyperbola, its key parameters (eccentricity and latus rectum), and the ability to manipulate trigonometric expressions and inequalities within a specified domain. We will break down the solution into clear, logical steps, explaining the rationale behind each.

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Step 1: Identify Parameters from the Standard Form of the Hyperbola

The first crucial step in any conic section problem is to correctly identify its type and extract its fundamental parameters by comparing it with the standard equation.

The given equation of the hyperbola is: $\frac{x^2}{\cos^2\theta} - \frac{y^2}{\sin^2\theta} = 1$

We compare this with the standard form of a hyperbola centered at the origin, with its transverse axis along the x-axis: $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$

By direct comparison, we can identify the squares of the semi-transverse axis ($a$) and the semi-conjugate axis ($b$): $a^2 = \cos^2\theta$ $b^2 = \sin^2\theta$

Now, we need to find $a$ and $b$. By definition, $a$ and $b$ represent lengths, so they must be positive values. $a = \sqrt{\cos^2\theta} = |\cos\theta|$ $b = \sqrt{\sin^2\theta} = |\sin\theta|$

The problem statement gives us a crucial domain for $\theta$: $0 < \theta < \frac{\pi}{2}$. In the first quadrant, both $\cos\theta$ and $\sin\theta$ are positive. Therefore, we can simplify: $a = \cos\theta$ $b = \sin\theta$

Why this step is important: Correctly identifying $a$ and $b$ is foundational. An error here would propagate through all subsequent calculations. The domain $0 < \theta < \frac{\pi}{2}$ is vital for removing the absolute value signs from $\sqrt{\cos^2\theta}$ and $\sqrt{\sin^2\theta}$.

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Step 2: Calculate Eccentricity and Apply the Given Condition

The eccentricity ($e$) is a characteristic parameter for a hyperbola that describes its shape, specifically how "open" its branches are. For a hyperbola of the form $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, the eccentricity is given by the formula: $e = \sqrt{1 + \frac{b^2}{a^2}}$ (Alternatively, $b^2 = a^2(e^2-1)$ can be used, which leads to the same formula for $e$.)

Now, we substitute the values of $a^2 = \cos^2\theta$ and $b^2 = \sin^2\theta$ that we found in Step 1: $e = \sqrt{1 + \frac{\sin^2\theta}{\cos^2\theta}}$

We recognize the fundamental trigonometric identity $\frac{\sin^2\theta}{\cos^2\theta} = \tan^2\theta$: $e = \sqrt{1 + \tan^2\theta}$

Another essential trigonometric identity states that $1 + \tan^2\theta = \sec^2\theta$. Substituting this into the expression for $e$: $e = \sqrt{\sec^2\theta}$ $e = |\sec\theta|$

Again, we refer to the given domain $0 < \theta < \frac{\pi}{2}$. In the first quadrant, $\cos\theta$ is positive, and therefore $\sec\theta = \frac{1}{\cos\theta}$ is also positive. So, we can simplify $|\sec\theta|$ to $\sec\theta$.

Thus, the eccentricity of the given hyperbola is: $e = \sec\theta$

The problem states that the eccentricity is greater than 2: $e > 2$ Substituting our expression for $e$: $\sec\theta > 2$

Why this step is important: This step converts the problem's condition into a trigonometric inequality involving $\theta$. Mastery of trigonometric identities is crucial here. Always remember to use the absolute value when taking the square root of a squared term, and then simplify using the given domain.

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Step 3: Determine the Valid Range for $\theta$

We need to solve the inequality $\sec\theta > 2$ within the domain $0 < \theta < \frac{\pi}{2}$. It's often easier to work with cosine when solving inequalities involving secant.

Recall that $\sec\theta = \frac{1}{\cos\theta}$. So the inequality becomes: $\frac{1}{\cos\theta} > 2$

Since $0 < \theta < \frac{\pi}{2}$, we know that $\cos\theta$ is positive (specifically, $0 < \cos\theta < 1$). Because $\cos\theta$ is positive, we can take the reciprocal of both sides of the inequality, but we must reverse the inequality sign: $\cos\theta < \frac{1}{2}$

Now, we need to find the values of $\theta$ in the interval $(0, \frac{\pi}{2})$ for which $\cos\theta < \frac{1}{2}$. We know the standard value $\cos\left(\frac{\pi}{3}\right) = \frac{1}{2}$.

In the first quadrant ($0 < \theta < \frac{\pi}{2}$), the cosine function is a strictly decreasing function. This means that as $\theta$ increases, $\cos\theta$ decreases. Therefore, for $\cos\theta < \frac{1}{2}$, $\theta$ must be greater than $\frac{\pi}{3}$.

Combining this with the initial domain $0 < \theta < \frac{\pi}{2}$, the valid range for $\theta$ that satisfies the eccentricity condition is: $\frac{\pi}{3} < \theta < \frac{\pi}{2}$

Why this step is important: This step establishes the critical domain for $\theta$ which will be used to determine the range of the latus rectum. A common mistake is forgetting to reverse the inequality sign when taking reciprocals, or misinterpreting the monotonicity of trigonometric functions in a given quadrant. Visualizing the unit circle or the graph of $\cos\theta$ can help avoid such errors.

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Step 4: Derive the Length of the Latus Rectum

The latus rectum (LR) is a chord of the hyperbola that passes through a focus and is perpendicular to the transverse axis. Its length is another key characteristic that helps define the shape and dimensions of the hyperbola. For a hyperbola of the form $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, the length of the latus rectum is given by the formula: $LR = \frac{2b^2}{a}$

Now, we substitute the values of $a = \cos\theta$ and $b^2 = \sin^2\theta$ that we found in Step 1: $LR = \frac{2\sin^2\theta}{\cos\theta}$

To make it easier to analyze this expression in the next step, we can rewrite it using trigonometric identities: $LR = 2 \left( \frac{\sin\theta}{\cos\theta} \right) \sin\theta$ $LR = 2 \tan\theta \sin\theta$

Why this step is important: This step translates the quantity we need to find (length of the latus rectum) into an expression solely in terms of $\theta$. This expression will then be analyzed over the determined range of $\theta$ to find its interval.

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Step 5: Find the Interval for the Length of the Latus Rectum

We need to find the range of the expression $LR = 2 \tan\theta \sin\theta$ for $\theta$ in the interval $\left( \frac{\pi}{3}, \frac{\pi}{2} \right)$. To do this, we will analyze the behavior of $\tan\theta$ and $\sin\theta$ within this specific interval.

Consider the interval $\left( \frac{\pi}{3}, \frac{\pi}{2} \right)$:

• Behavior of $\tan\theta$:

  • In the first quadrant, $\tan\theta$ is a strictly increasing function.
  • As $\theta$ approaches $\frac{\pi}{3}$ from the right ($\theta \to \frac{\pi}{3}^+$), $\tan\theta$ approaches $\tan\left(\frac{\pi}{3}\right) = \sqrt{3}$.
  • As $\theta$ approaches $\frac{\pi}{2}$ from the left ($\theta \to \frac{\pi}{2}^-$), $\tan\theta$ approaches $\infty$.
  • So, $\tan\theta \in (\sqrt{3}, \infty)$.

• Behavior of $\sin\theta$:

  • In the first quadrant, $\sin\theta$ is also a strictly increasing function.
  • As $\theta$ approaches $\frac{\pi}{3}$ from the right ($\theta \to \frac{\pi}{3}^+$), $\sin\theta$ approaches $\sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}$.
  • As $\theta$ approaches $\frac{\pi}{2}$ from the left ($\theta \to \frac{\pi}{2}^-$), $\sin\theta$ approaches $\sin\left(\frac{\pi}{2}\right) = 1$.
  • So, $\sin\theta \in \left( \frac{\sqrt{3}}{2}, 1 \right)$.

Since both $\tan\theta$ and $\sin\theta$ are positive and strictly increasing functions in the interval $\left( \frac{\pi}{3}, \frac{\pi}{2} \right)$, their product, $LR = 2 \tan\theta \sin\theta$, will also be a strictly increasing function in this interval.

Therefore, the minimum value of $LR$ will be approached as $\theta$ approaches $\frac{\pi}{3}$ from the right, and the maximum value will be approached as $\theta$ approaches $\frac{\pi}{2}$ from the left.

Lower bound for LR: As $\theta \to \frac{\pi}{3}^+$, $LR_{lower\_bound} = 2 \times \lim_{\theta \to \frac{\pi}{3}^+} (\tan\theta) \times \lim_{\theta \to \frac{\pi}{3}^+} (\sin\theta)$ $LR_{lower\_bound} = 2 \times \tan\left(\frac{\pi}{3}\right) \times \sin\left(\frac{\pi}{3}\right)$ $LR_{lower\_bound} = 2 \times \sqrt{3} \times \frac{\sqrt{3}}{2}$ $LR_{lower\_bound} = 2 \times \frac{3}{2}$ $LR_{lower\_bound} = 3$ Since $\theta > \frac{\pi}{3}$, this value $LR=3$ is not included in the interval. It defines the open lower bound.

Upper bound for LR: As $\theta \to \frac{\pi}{2}^-$, $LR_{upper\_bound} = 2 \times \lim_{\theta \to \frac{\pi}{2}^-} (\tan\theta) \times \lim_{\theta \to \frac{\pi}{2}^-} (\sin\theta)$ $LR_{upper\_bound} = 2 \times (\infty) \times 1$ $LR_{upper\_bound} = \infty$ Since $\theta < \frac{\pi}{2}$, this upper bound is also not included.

Combining these results, the length of the latus rectum $LR$ lies in the open interval: $(3, \infty)$

Why this step is important: This is the culmination of all previous steps. Analyzing the monotonicity (increasing or decreasing nature) of the function for $LR$ over the derived range of $\theta$ is crucial. Remember that for open intervals of $\theta$, the corresponding range of the function will also be open, meaning the endpoints are not included.

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Summary and Key Takeaway:

This problem is an excellent example of how JEE questions often integrate multiple mathematical concepts. We started by correctly