For some θ ∈ (0, π/2), let the eccentricity and the length of the latus rectum of the hyperbola x² -

Q: Wait, why did you switch the standard form of the ellipse? How do I know if the major axis is on the x or y axis?

Great catch! In the ellipse equation $x^2 \sec^2 \theta + y^2 = 6$, we rewrite it as $\frac{x^2}{6 \cos^2 \theta} + \frac{y^2}{6} = 1$. Since $\theta \in (0, \frac{\pi}{2})$, we know $\cos^2 \theta < 1$, which means $6 \cos^2 \theta < 6$. Because the denominator under $y^2$ is larger, the major axis must be along the y-axis. Always compare the denominators to identify the major axis before calculating eccentricity!

Q: Why can't I just use $e^2 = 1 - \frac{b^2}{a^2}$ for the ellipse? It feels like I'm always getting the sign wrong.

The formula $e^2 = 1 - \frac{\text{minor}^2}{\text{major}^2}$ is the safest way to remember it. By identifying the major axis first, you ensure you are always dividing the smaller value by the larger value. This keeps $e^2$ between $0$ and $1$, which is exactly what we expect for an ellipse.

Q: The quadratic equation $2 \cos^4 \theta + \cos^2 \theta - 1 = 0$ looks scary. Did I factor it correctly?

You absolutely did! Think of it as $2u^2 + u - 1 = 0$ where $u = \cos^2 \theta$. Factoring this gives $(2u - 1)(u + 1) = 0$. Since $\cos^2 \theta$ cannot be $-1$, we discard that root. You are on the right track—trust your algebraic instincts!

Q: Is there a shortcut to calculating the final expression $(\frac{l_1 l_2}{e_1 e_2}) \tan^2 \theta$?

While you could try to simplify the expression algebraically before plugging in $\theta$, it is often safer and faster to find the specific value of $\theta$ first. Once you know $\cos^2 \theta = 1/2$, the values of $e$ and $l$ become simple numbers, making the final arithmetic much less prone to error.

Q: Why is the eccentricity of the hyperbola $e_1^2 = 1 + \cos^2 \theta$ and not something else?

For the hyperbola $\frac{x^2}{8} - \frac{y^2}{8 \cos^2 \theta} = 1$, we have $a^2 = 8$ and $b^2 = 8 \cos^2 \theta$. The definition of eccentricity for a hyperbola is $e^2 = 1 + \frac{b^2}{a^2}$. Substituting our values, we get $1 + \frac{8 \cos^2 \theta}{8}$, which simplifies beautifully to $1 + \cos^2 \theta$. It's all about staying organized with your $a^2$ and $b^2$ values!

Question

For some $	heta \in (0, \frac{\pi}{2})$, let the eccentricity and the length of the latus rectum of the hyperbola $x^2 - y^2 \sec^2 	heta = 8$ be $e_1$ and $l_1$, respectively, and let the eccentricity and the length of the latus rectum of the ellipse $x^2 \sec^2 	heta + y^2 = 6$ be $e_2$ and $l_2$, respectively. If $e_1^2 = e_2^2 (\sec^2 	heta + 1)$, then $(\frac{l_1 l_2}{e_1 e_2}) 	an^2 	heta$ is equal to .........

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler of the mathematical landscape. Today, we are not just solving a problem; we are choreographing a dance between two of the most beautiful shapes in geometry: the hyperbola and the ellipse.

Beneath the surface of $\sec^2 	heta$ and $e^2$, there is a hidden symmetry waiting to be revealed.

### Bringing Order to Chaos

Our journey begins by standardizing our subjects. A hyperbola is defined by its standard form, $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.

Looking at our given equation, $x^2 - y^2 \sec^2 	heta = 8$, we divide by $8$ to find:

$$ \frac{x^2}{8} - \frac{y^2}{8 \cos^2 	heta} = 1 $$

Suddenly, the fog clears. We identify $a^2 = 8$ and $b^2 = 8 \cos^2 	heta$.

For the hyperbola, the eccentricity $e_1$ is governed by the relationship $e_1^2 = 1 + \frac{b^2}{a^2}$. Substituting our values, we find:

$$ e_1^2 = 1 + \frac{8 \cos^2 	heta}{8} = 1 + \cos^2 	heta $$

The latus rectum, that elegant chord passing through the focus, is given by $l_1 = \frac{2b^2}{a}$. With $a = \sqrt{8} = 2\sqrt{2}$, we get:

$$ l_1 = \frac{2(8 \cos^2 	heta)}{2\sqrt{2}} = 4\sqrt{2} \cos^2 	heta $$

### The Ellipse's Secret

Next, we turn our attention to the ellipse: $x^2 \sec^2 	heta + y^2 = 6$. Dividing by $6$, we get:

$$ \frac{x^2}{6 \cos^2 	heta} + \frac{y^2}{6} = 1 $$

Here is where many students stumble. Because $\cos^2 	heta < 1$, the denominator under $y^2$ is larger, which tells us the major axis is vertical.

For this ellipse, $a^2 = 6 \cos^2 	heta$ and $b^2 = 6$. The eccentricity $e_2$ follows:

$$ e_2^2 = 1 - \frac{a^2}{b^2} = 1 - \frac{6 \cos^2 	heta}{6} = 1 - \cos^2 	heta = \sin^2 	heta $$

The latus rectum is calculated as:

$$ l_2 = \frac{2a^2}{b} = \frac{2(6 \cos^2 	heta)}{\sqrt{6}} = 2\sqrt{6} \cos^2 	heta $$

### The Convergence

Now, we apply the master constraint: $e_1^2 = e_2^2 (\sec^2 	heta + 1)$. Substituting our expressions, we have:

$$ 1 + \cos^2 	heta = \sin^2 	heta \left( \frac{1}{\cos^2 	heta} + 1 ight) $$

Expanding the right side, we get $\sin^2 	heta + \frac{\sin^2 	heta}{\cos^2 	heta} = \sin^2 	heta + 	an^2 	heta$.

This is the moment of truth. We replace $\sin^2 	heta$ with $1 - \cos^2 	heta$ to get:

$$ 1 + \cos^2 	heta = 1 - \cos^2 	heta + 	an^2 	heta $$

Simplifying this yields $2 \cos^2 	heta = 	an^2 	heta$. By rewriting $	an^2 	heta$ as $\frac{1 - \cos^2 	heta}{\cos^2 	heta}$, we arrive at the quadratic:

$$ 2 \cos^4 	heta + \cos^2 	heta - 1 = 0 $$

Factoring this, we find $(2 \cos^2 	heta - 1)(\cos^2 	heta + 1) = 0$. Since $\cos^2 	heta$ must be positive, we conclude $\cos^2 	heta = \frac{1}{2}$, which means $	heta = \frac{\pi}{4}$.

### The Grand Finale

With $	heta = \frac{\pi}{4}$, the world simplifies. We find $e_1 = \sqrt{3/2}$, $e_2 = 1/\sqrt{2}$, $l_1 = 2\sqrt{2}$, and $l_2 = \sqrt{6}$.

Plugging these into our final expression $\left( \frac{l_1 l_2}{e_1 e_2} ight) 	an^2 	heta$, we calculate:

$$ \frac{(2\sqrt{2})(\sqrt{6})}{(\sqrt{3/2})(1/\sqrt{2})} 	imes 1 $$

The numerator becomes $2\sqrt{12} = 4\sqrt{3}$, and the denominator simplifies to $\frac{\sqrt{3}}{2}$.

Dividing $4\sqrt{3}$ by $\frac{\sqrt{3}}{2}$ gives us exactly **8**.

Enter Numerical Value:

Visualized Solution

Hyperbola Standard Form

Hyperbola Parameters $e_{1}$ and $l_{1}$

Ellipse Standard Form

Ellipse Parameters $e_{2}$ and $l_{2}$

Applying the Given Condition

Simplifying the Equation

Solving for $θ$

Finding $cos^{2} θ$

Evaluating Parameters

Final Calculation

The Sigma Insight: Standard Equations of Parabola, Ellipse, and Hyperbola

Analyzing the Setup

Bringing Order to Chaos

The Ellipse's Secret

The Convergence

The Grand Finale

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