This problem is a comprehensive test of your understanding of the fundamental properties of hyperbolas, specifically distinguishing between a standard hyperbola and a conjugate hyperbola. We will systematically break down the problem by first identifying the type of each hyperbola and then applying the relevant formulas for eccentricity, latus rectum, and transverse axis length to determine the unknown parameters.

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Key Concepts and Formulas for Hyperbolas

A hyperbola's equation dictates its orientation and the specific formulas for its properties. It's crucial to correctly identify the "semi-transverse axis" (the denominator under the positive term) and the "semi-conjugate axis" for each type.

1. Standard Hyperbola ($\mathrm{H}_1$):

   • Equation: $\frac{x^2}{\mathrm{a}^2}-\frac{y^2}{\mathrm{~b}^2}=1$.
   • Orientation: Its transverse axis lies along the x-axis. The term with $x^2$ is positive.
   • Semi-transverse axis length: $\mathrm{a}$ (along the x-axis).
   • Semi-conjugate axis length: $\mathrm{b}$ (along the y-axis).
   • Length of Transverse axis: $2\mathrm{a}$. This is the distance between the two vertices.
   • Length of Conjugate axis: $2\mathrm{b}$.
   • Eccentricity ($e_1$): The relationship between $\mathrm{a}$, $\mathrm{b}$, and $e_1$ is given by $e_1^2 = 1 + \frac{\mathrm{b}^2}{\mathrm{a}^2}$. This formula arises from the fundamental relationship $c^2 = a^2 + b^2$ (where $c$ is the distance from the center to a focus) and the definition $e = c/a$.
   • Length of Latus Rectum (L.R.): $\frac{2\mathrm{b}^2}{\mathrm{a}}$. This is the length of a chord passing through a focus and perpendicular to the transverse axis.

2. Conjugate Hyperbola ($\mathrm{H}_2$):

   • Equation: $-\frac{x^2}{\mathrm{~A}^2}+\frac{y^2}{\mathrm{~B}^2}=1$, which is often rewritten as $\frac{y^2}{\mathrm{~B}^2}-\frac{x^2}{\mathrm{~A}^2}=1$.
   • Orientation: Its transverse axis lies along the y-axis. The term with $y^2$ is positive.
   • Semi-transverse axis length: $\mathrm{B}$ (along the y-axis).
   • Semi-conjugate axis length: $\mathrm{A}$ (along the x-axis).
   • Length of Transverse axis: $2\mathrm{B}$. This is the distance between the two vertices.
   • Length of Conjugate axis: $2\mathrm{A}$.
   • Eccentricity ($e_2$): The relationship between $\mathrm{A}$, $\mathrm{B}$, and $e_2$ is given by $e_2^2 = 1 + \frac{\mathrm{A}^2}{\mathrm{~B}^2}$. Notice the roles of the semi-axes are swapped compared to the standard hyperbola, reflecting the change in orientation.
   • Length of Latus Rectum (L.R.): $\frac{2\mathrm{A}^2}{\mathrm{~B}}$.

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Step-by-Step Solution

We will analyze each hyperbola individually, using the provided information to determine its parameters.

1. Analyzing Hyperbola $\mathrm{H}_1$

The equation for $\mathrm{H}_1$ is given as $\frac{x^2}{\mathrm{a}^2}-\frac{y^2}{\mathrm{~b}^2}=1$. This is a standard hyperbola with its transverse axis along the x-axis.

• Given Information for $\mathrm{H}_1$:
  • Length of latus rectum: $15\sqrt{2}$
  • Eccentricity $e_1 = \sqrt{\frac{5}{2}}$

Step 1.1: Using Eccentricity to Relate $\mathrm{a}$ and $\mathrm{b}$

• Concept: The eccentricity formula connects the lengths of the semi-transverse and semi-conjugate axes.
• Formula: For $\mathrm{H}_1$, $e_1^2 = 1 + \frac{\mathrm{b}^2}{\mathrm{a}^2}$.
• Application: We are given $e_1 = \sqrt{\frac{5}{2}}$, so $e_1^2 = \frac{5}{2}$. Substitute this value into the eccentricity formula: $\frac{5}{2} = 1 + \frac{\mathrm{b}^2}{\mathrm{a}^2}$
• Explanation: Our goal is to find a relationship between $\mathrm{a}^2$ and $\mathrm{b}^2$. We isolate the ratio $\frac{\mathrm{b}^2}{\mathrm{a}^2}$: $\frac{\mathrm{b}^2}{\mathrm{a}^2} = \frac{5}{2} - 1 = \frac{5}{2} - \frac{2}{2} = \frac{3}{2}$ This gives us a crucial relationship: $\mathrm{b}^2 = \frac{3}{2}\mathrm{a}^2 \quad \text{(Equation 1.1)}$

Step 1.2: Using Length of Latus Rectum to Find 'a' and 'b'

• Concept: The latus rectum length provides another equation relating 'a' and 'b'. Combining this with the eccentricity relationship allows us to solve for 'a' and 'b' uniquely.

• Formula: For $\mathrm{H}_1$, the length of the latus rectum is $\frac{2\mathrm{b}^2}{\mathrm{a}}$.

• Application: We are given that this length is $15\sqrt{2}$: $\frac{2\mathrm{b}^2}{\mathrm{a}} = 15\sqrt{2}$

• Explanation: To solve for 'a' and 'b', we substitute the expression for $\mathrm{b}^2$ from Equation 1.1 into this equation. This eliminates 'b' and leaves an equation solely in terms of 'a': $\frac{2 \left(\frac{3}{2}\mathrm{a}^2\right)}{\mathrm{a}} = 15\sqrt{2}$ Simplify the expression: $\frac{3\mathrm{a}^2}{\mathrm{a}} = 15\sqrt{2}$ $3\mathrm{a} = 15\sqrt{2}$ Divide by 3 to find the value of 'a': $\mathrm{a} = \frac{15\sqrt{2}}{3} = 5\sqrt{2}$

• Explanation: Now that we have the value of 'a', we can find $\mathrm{b}^2$ using Equation 1.1: $\mathrm{b}^2 = \frac{3}{2}\mathrm{a}^2 = \frac{3}{2}(5\sqrt{2})^2$ Calculate $(5\sqrt{2})^2$: $(5\sqrt{2})^2 = 5^2 \times (\sqrt{2})^2 = 25 \times 2 = 50$. $\mathrm{b}^2 = \frac{3}{2}(50) = 3 \times 25 = 75$ From this, we can also find $\mathrm{b} = \sqrt{75} = 5\sqrt{3}$.

• Summary for $\mathrm{H}_1$:

  • Semi-transverse axis: $\mathrm{a} = 5\sqrt{2}$
  • Semi-conjugate axis: $\mathrm{b} = 5\sqrt{3}$
  • Length of transverse axis: $2\mathrm{a} = 2(5\sqrt{2}) = 10\sqrt{2}$. This value will be used in the next step.

2. Analyzing Hyperbola $\mathrm{H}_2$

The equation for $\mathrm{H}_2$ is given as $-\frac{x^2}{\mathrm{~A}^2}+\frac{y^2}{\mathrm{~B}^2}=1$. This is a conjugate hyperbola with its transverse axis along the y-axis.

• Given Information for $\mathrm{H}_2$:
  • Length of latus rectum: $12\sqrt{5}$
  • Eccentricity $e_2$ (which we need to find)
  • Product of the lengths of their transverse axes: $100\sqrt{10}$

Step 2.1: Using the Product of Transverse Axes to Find 'B'

• Concept: The product of transverse axes lengths provides a direct link between $\mathrm{H}_1$ and $\mathrm{H}_2$, allowing us to determine a parameter for $\mathrm{H}_2$.
• Formula:
  • Length of transverse axis for $\mathrm{H}_1$ is $2\mathrm{a}$.
  • Length of transverse axis for $\mathrm{H}_2$ is $2\mathrm{B}$.
• Application: We are given that their product is $100\sqrt{10}$: $(2\mathrm{a}) \times (2\mathrm{B}) = 100\sqrt{10}$
• Explanation: We already found $2\mathrm{a} = 10\sqrt{2}$ from our analysis of $\mathrm{H}_1$. Substitute this value: $(10\sqrt{2}) \times (2\mathrm{B}) = 100\sqrt{10}$ $20\sqrt{2}\mathrm{B} = 100\sqrt{10}$
• Explanation: To find $\mathrm{B}$, divide both sides by $20\sqrt{2}$: $\mathrm{B} = \frac{100\sqrt{10}}{20\sqrt{2}}$ Simplify the numerical and radical parts separately: $\mathrm{B} = \left(\frac{100}{20}\right) \times \left(\frac{\sqrt{10}}{\sqrt{2}}\right) = 5 \times \sqrt{\frac{10}{2}} = 5\sqrt{5}$ So, for $\mathrm{H}_2$, the semi-transverse axis length is $\mathrm{B} = 5\sqrt{5}$. We will also need $\mathrm{B}^2$: $\mathrm{B}^2 = (5\sqrt{5})^2 = 25 \times 5 = 125$.

Step 2.2: Using Length of Latus Rectum to Find 'A'

• Concept: The latus rectum formula for a conjugate hyperbola connects its semi-conjugate axis length ($\mathrm{A}$) and semi-transverse axis length ($\mathrm{B}$).

• Formula: For $\mathrm{H}_2$, the length of the latus rectum is $\frac{2\mathrm{A}^2}{\mathrm{~B}}$.

• Application: We are given that this length is $12\sqrt{5}$: $\frac{2\mathrm{A}^2}{\mathrm{~B}} = 12\sqrt{5}$

• Explanation: Substitute the value of $\mathrm{B} = 5\sqrt{5}$ that we just found into this equation: $\frac{2\mathrm{A}^2}{5\sqrt{5}} = 12\sqrt{5}$

• Explanation: To find $\mathrm{A}^2$, multiply both sides by $5\sqrt{5}$: $2\mathrm{A}^2 = 12\sqrt{5} \times 5\sqrt{5}$ $2\mathrm{A}^2 = 12 \times 5 \times (\sqrt{5} \times \sqrt{5})$ $2\mathrm{A}^2 = 60 \times 5$ $2\mathrm{A}^2 = 300$ Divide by 2 to find $\mathrm{A}^2$: $\mathrm{A}^2 = 150$ (We could find $\mathrm{A} = \sqrt{150} = 5\sqrt{6}$, but $\mathrm{A}^2$ is sufficient for calculating eccentricity).

• Summary for $\mathrm{H}_2$:

  • Semi-transverse axis squared: $\mathrm{B}^2 = 125$
  • Semi-conjugate axis squared: $\mathrm{A}^2 = 150$

3. Calculating Eccentricity $\mathrm{e}_2$ for $\mathrm{H}_2$

Now that we have $\mathrm{A}^2$ and $\mathrm{B}^2$ for $\mathrm{H}_2$, we can calculate its eccentricity $e_2$.

• Concept: The eccentricity formula directly uses the squares of the semi-transverse and semi-conjugate axis lengths.
• Formula: For $\mathrm{H}_2$, the square of the eccentricity is $e_2^2 = 1 + \frac{\mathrm{A}^2}{\mathrm{~B}^2}$.
• Application: Substitute the values $\mathrm{A}^2 = 150$ and $\mathrm{B}^2 = 125$: $e_2^2 = 1 + \frac{150}{125}$
• Explanation: Simplify the fraction $\frac{150}{12