Here's a detailed and educational rewrite of the solution:

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1. Fundamental Concepts & Formulas for Conic Sections

To tackle this problem, we need a solid understanding of the standard forms and key properties of hyperbolas and ellipses, especially regarding their eccentricity and latus rectum.

• Hyperbola:

  • Standard Form (transverse axis along x-axis): The equation is given by $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$. Here, $a$ is the length of the semi-transverse axis, and $b$ is the length of the semi-conjugate axis.
  • Eccentricity ($e_H$): This value quantifies the "openness" of the hyperbola. It's always greater than 1 ($e_H > 1$). The formula is $e_H = \sqrt{1 + \frac{b^2}{a^2}}$.

• Ellipse:

  • Standard Form: The general equation is $\frac{x^2}{A^2} + \frac{y^2}{B^2} = 1$. Here, $A$ and $B$ represent the lengths related to the semi-axes.
  • Determining Major and Minor Axes: This is crucial for correctly applying eccentricity and latus rectum formulas.
    • If $A^2 > B^2$, the major axis is along the x-axis. The semi-major axis is $A$, and the semi-minor axis is $B$.
    • If $B^2 > A^2$, the major axis is along the y-axis. The semi-major axis is $B$, and the semi-minor axis is $A$.
  • Eccentricity ($e_E$): This value describes how "flattened" the ellipse is. It's always between 0 and 1 ($0 < e_E < 1$).
    • If the major axis is along the x-axis ($A^2 > B^2$): $e_E = \sqrt{1 - \frac{B^2}{A^2}}$
    • If the major axis is along the y-axis ($B^2 > A^2$): $e_E = \sqrt{1 - \frac{A^2}{B^2}}$
  • Length of Latus Rectum ($L_E$): This is the length of the chord passing through a focus and perpendicular to the major axis.
    • If the major axis is along the x-axis ($A^2 > B^2$): $L_E = \frac{2B^2}{A}$
    • If the major axis is along the y-axis ($B^2 > A^2$): $L_E = \frac{2A^2}{B}$

Tip for Ellipse Parameters: Always identify $A^2$ and $B^2$ first, then determine which is larger to correctly assign the semi-major and semi-minor axes, and consequently, the correct formulas for eccentricity and latus rectum.

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2. Decoding the Hyperbola's Properties

Our first task is to transform the given hyperbola equation into its standard form to extract its parameters and calculate its eccentricity.

Step 2.1: Standardizing the Hyperbola Equation The given equation for the hyperbola is: $x^2 - y^2 \sec^2 \theta = 10$ Why this step? The standard form of a hyperbola has '1' on the right-hand side. To achieve this, we divide the entire equation by 10. $\frac{x^2}{10} - \frac{y^2 \sec^2 \theta}{10} = 1$ Why rearrange the second term? To match the standard form $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, the $y^2$ term should have a coefficient of 1, with any multipliers incorporated into its denominator. $\frac{x^2}{10} - \frac{y^2}{10/\sec^2 \theta} = 1$ Now, we use the trigonometric identity $\frac{1}{\sec^2 \theta} = \cos^2 \theta$ to simplify the denominator: $\frac{x^2}{10} - \frac{y^2}{10 \cos^2 \theta} = 1$

Step 2.2: Identifying Hyperbola Parameters ($a_H^2$ and $b_H^2$) By comparing this standard form with $\frac{x^2}{a_H^2} - \frac{y^2}{b_H^2} = 1$, we can directly identify the squares of the semi-transverse and semi-conjugate axes: $a_H^2 = 10$ $b_H^2 = 10 \cos^2 \theta$

Step 2.3: Calculating Hyperbola Eccentricity ($e_H$) Using the formula for hyperbola eccentricity, $e_H = \sqrt{1 + \frac{b_H^2}{a_H^2}}$: $e_H = \sqrt{1 + \frac{10 \cos^2 \theta}{10}}$ Why simplify? The '10' in the numerator and denominator cancels out, simplifying the expression. $e_H = \sqrt{1 + \cos^2 \theta} \quad \text{(Equation 1)}$ This gives us the eccentricity of the hyperbola in terms of $\theta$.

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3. Unraveling the Ellipse's Characteristics

We apply a similar methodical approach to the ellipse: standardize its equation, identify its parameters, and calculate its eccentricity.

Step 3.1: Standardizing the Ellipse Equation The given equation for the ellipse is: $x^2 \sec^2 \theta + y^2 = 5$ Why this step? Like the hyperbola, the standard form of an ellipse requires '1' on the right-hand side. We divide the entire equation by 5. $\frac{x^2 \sec^2 \theta}{5} + \frac{y^2}{5} = 1$ Why rearrange the first term? To match the standard form $\frac{x^2}{A^2} + \frac{y^2}{B^2} = 1$, the $x^2$ term's coefficient needs to be incorporated into its denominator. $\frac{x^2}{5/\sec^2 \theta} + \frac{y^2}{5} = 1$ Using the identity $\frac{1}{\sec^2 \theta} = \cos^2 \theta$, the equation becomes: $\frac{x^2}{5 \cos^2 \theta} + \frac{y^2}{5} = 1$

Step 3.2: Identifying Ellipse Parameters ($A^2$ and $B^2$) and Determining Major/Minor Axes By comparing this with the standard form $\frac{x^2}{A^2} + \frac{y^2}{B^2} = 1$, we identify: $A^2 = 5 \cos^2 \theta$ $B^2 = 5$ Why determine major/minor axes? This is a critical step for ellipses. The choice of eccentricity and latus rectum formulas depends entirely on whether the major axis is horizontal or vertical. We need to compare $A^2$ and $B^2$. We are given that $\theta \in \left( {0,{\pi \over 2}} \right)$. For this range of $\theta$:

• $\cos \theta$ is a positive value between 0 and 1 (exclusive, i.e., $0 < \cos \theta < 1$).
• Therefore, $\cos^2 \theta$ is also a positive value between 0 and 1 (exclusive, i.e., $0 < \cos^2 \theta < 1$).
• This implies $A^2 = 5 \cos^2 \theta$ will be less than $5 \times 1 = 5$. So, we have $A^2 < B^2$. Since $B^2 > A^2$, the major axis of the ellipse is along the y-axis.
• The semi-major axis is $B = \sqrt{B^2} = \sqrt{5}$.
• The semi-minor axis is $A = \sqrt{A^2} = \sqrt{5 \cos^2 \theta}$.

Common Mistake: Assuming the major axis is always associated with the $x^2$ term or simply picking the larger denominator without considering the range of $\theta$. Always analyze the values of $A^2$ and $B^2$ to correctly determine the major axis.

Step 3.3: Calculating Ellipse Eccentricity ($e_E$) Since the major axis is along the y-axis, we use the formula $e_E = \sqrt{1 - \frac{A^2}{B^2}}$: $e_E = \sqrt{1 - \frac{5 \cos^2 \theta}{5}}$ Why simplify? The '5' in the numerator and denominator cancels out. $e_E = \sqrt{1 - \cos^2 \theta}$ Using the fundamental trigonometric identity $\sin^2 \theta + \cos^2 \theta = 1$, which means $1 - \cos^2 \theta = \sin^2 \theta$: $e_E = \sqrt{\sin^2 \theta}$ Why is it $\sin \theta$ and not $|\sin \theta|$? We are given that $\theta \