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Q: Wait, why does the inequality sign flip when I go from $\sec \theta > 2$ to $\cos \theta < \frac{1}{2}$?

Great catch! This is a classic trap. Remember that the function $f(x) = \frac{1}{x}$ is a strictly decreasing function for positive $x$. Since $\sec \theta$ and $2$ are both positive in the first quadrant, taking the reciprocal reverses the inequality. Always be vigilant with inequalities involving trigonometric functions!

Q: Why is the interval $(3, \infty)$ and not $[3, \infty)$?

Excellent question. Look at the constraint $0 2$ implies $\theta > \frac{\pi}{3}$. Since the inequality is strict ($>$), $\theta$ can never actually equal $\frac{\pi}{3}$. Therefore, the value of the latus rectum can approach 3 but never reach it. Hence, we use the open interval.

Q: Can I use derivatives to find the range of the latus rectum?

Absolutely! You can define $f(\theta) = 2(\sec \theta - \cos \theta)$ and find $f'(\theta)$. You will see that $f'(\theta) = 2(\sec \theta \tan \theta + \sin \theta)$, which is positive for $\theta \in (\frac{\pi}{3}, \frac{\pi}{2})$. This confirms the function is strictly increasing. However, observing the behavior of $\sec \theta$ and $\cos \theta$ directly is often faster in an exam setting.

Question

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

Options:

(A) (2, 3]

(B) (3, \infty)

(C) (3/2, 2]

(D) (1, 3/2]

Sigma Percentile · Accepted Answer

### Analyzing the Setup We are presented with the hyperbola equation: $$\frac{x^2}{\cos^2 heta} - \frac{y^2}{\sin^2 heta} = 1$$ By comparing this to the standard form $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, we identify the parameters as $a = \cos heta$ and $b = \sin heta$. As $ heta$ varies, the geometry of this hyperbola evolves. Our goal is to determine the behavior of its eccentricity and latus rectum under the given constraints. ### The Eccentricity Dance The eccentricity $e$ of a hyperbola is defined by the relationship: $$e = \sqrt{1 + \frac{b^2}{a^2}}$$ Substituting our specific parameters $a$ and $b$ into this formula, we obtain: $$e = \sqrt{1 + \frac{\sin^2 heta}{\cos^2 heta}}$$ Since $\frac{\sin^2 heta}{\cos^2 heta} = an^2 heta$, the expression simplifies elegantly to: $$e = \sqrt{1 + an^2 heta} = \sec heta$$ ### The Inequality Trap The problem imposes the constraint $e > 2$, which implies $\sec heta > 2$. Given that we are working in the first quadrant where $0 < heta < \frac{\pi}{2}$, we must invert this to find the range for $\cos heta$. Recalling that the inequality sign flips when taking the reciprocal of positive values, we get: $$\cos heta < \frac{1}{2}$$ Since $\cos heta = \frac{1}{2}$ at $ heta = \frac{\pi}{3}$ and the cosine function is strictly decreasing in the first quadrant, the condition $\cos heta < \frac{1}{2}$ corresponds to the interval $ heta \in (\frac{\pi}{3}, \frac{\pi}{2})$. ### The Latus Rectum Journey The length of the latus rectum ($L.R.$) for a hyperbola is given by the formula: $$L.R. = \frac{2b^2}{a}$$ Substituting our values for $a$ and $b$, we have: $$L.R. = \frac{2\sin^2 heta}{\cos heta}$$ Using the trigonometric identity $\sin^2 heta = 1 - \cos^2 heta$, we can rewrite the expression as: $$L.R. = \frac{2(1 - \cos^2 heta)}{\cos heta} = 2(\sec heta - \cos heta)$$ ### Final Calculation Let $f( heta) = 2(\sec heta - \cos heta)$. We analyze the behavior of this function as $ heta$ increases from $\frac{\pi}{3}$ to $\frac{\pi}{2}$. As $ heta$ approaches $\frac{\pi}{3}$, the value of the latus rectum approaches: $$L.R. o 2\left(2 - \frac{1}{2} ight) = 2\left(\frac{3}{2} ight) = 3$$ As $ heta$ approaches $\frac{\pi}{2}$, the term $\sec heta$ approaches $\infty$ while $\cos heta$ approaches $0$. Consequently, the latus rectum approaches $\infty$. The range of the length of the latus rectum is therefore **$(3, \infty)$**.

Animated Solution for Mathematics - Conic Sections: Let $0 < θ < \frac{π}{2}$ . If the eccentricity of the hyperbola $\frac{x ^{2}}{c o s ^{2} θ} - \frac{y ^{2}}{s i n ^{2} θ} = 1$ is greater than 2, then the length of its latus rectum lies in the interval :

Select Answer:

Visualized Solution

Analysis of the Hyperbola Equation

The Eccentricity Formula

Simplifying Eccentricity $e$

Applying the Constraint $e > 2$

Determining the Range of $θ$

Latus Rectum Formula

Substituting $a$ and $b$ in $L . R .$

Simplifying the $L . R .$ Expression

Analyzing the Function $f (θ)$

Calculating the Lower Bound

Calculating the Upper Bound

Final Conclusion

The Sigma Insight: Foci, Directrices, and Eccentricity

Analyzing the Setup

The Eccentricity Dance

The Inequality Trap

The Latus Rectum Journey

Final Calculation

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