This problem is an excellent test of your understanding of the fundamental properties of conic sections, specifically ellipses and hyperbolas. It requires you to accurately identify parameters, calculate eccentricity and foci for one conic, and then use those derived properties along with given conditions to determine the characteristics of another conic.

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1. Understanding the Problem and Key Conic Section Properties

Our goal is to find the value of $3\alpha^2 + 2\beta^2$, where $\alpha$ is the length of the transverse axis and $\beta$ is the length of the conjugate axis of a hyperbola $H$. We are given an ellipse $E$ and specific relationships between its properties and those of hyperbola $H$.

To successfully solve this, we need to recall the essential definitions and formulas for ellipses and hyperbolas centered at $(h,k)$.

• Ellipse (Horizontal Major Axis): For an ellipse with the standard equation: $\frac{(x-h)^2}{a_E^2} + \frac{(y-k)^2}{b_E^2} = 1 \quad (\text{where } a_E > b_E)$

  • Center: $(h,k)$
  • Semi-major axis length: $a_E$
  • Semi-minor axis length: $b_E$
  • Eccentricity: $e_E = \sqrt{1 - \frac{b_E^2}{a_E^2}}$
  • Distance from center to focus: $c_E = a_E e_E$
  • Foci: $(h \pm c_E, k)$

• Hyperbola (Horizontal Transverse Axis): For a hyperbola with the standard equation: $\frac{(x-h)^2}{a_H^2} - \frac{(y-k)^2}{b_H^2} = 1$

  • Center: $(h,k)$
  • Semi-transverse axis length: $a_H$
  • Semi-conjugate axis length: $b_H$
  • Eccentricity: $e_H = \sqrt{1 + \frac{b_H^2}{a_H^2}}$
  • Distance from center to focus: $c_H = a_H e_H$
  • Fundamental Relationship: $c_H^2 = a_H^2 + b_H^2$ (This equation directly links $a_H, b_H, c_H$ for a hyperbola).
  • Foci: $(h \pm c_H, k)$

• Important Note on Notation: To avoid confusion, we will consistently use subscripts 'E' for ellipse parameters ($a_E, b_E, e_E, c_E$) and 'H' for hyperbola parameters ($a_H, b_H, e_H, c_H$).

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2. Analyzing the Ellipse $E$

The given equation of the ellipse is: $E: \frac{(x-1)^2}{100}+\frac{(y-1)^2}{75}=1$

2.1 Identify the Center and Semi-Axes of Ellipse $E$

• WHY: To establish the fundamental dimensions and location of the ellipse, which are essential for calculating its eccentricity and foci.
• Concept: We compare the given equation with the standard form $\frac{(x-h)^2}{a_E^2} + \frac{(y-k)^2}{b_E^2} = 1$.
• Step-by-step working:
  • By direct comparison, the center of the ellipse $(h_E, k_E)$ is $(1,1)$.
  • The term under $(x-1)^2$ is $100$, so $a_E^2 = 100$.
    • This implies $a_E = \sqrt{100} = 10$. This is the length of the semi-major axis.
  • The term under $(y-1)^2$ is $75$, so $b_E^2 = 75$.
    • This implies $b_E = \sqrt{75} = \sqrt{25 \times 3} = 5\sqrt{3}$. This is the length of the semi-minor axis.
• Tip: Since $a_E^2 > b_E^2$ and $a_E^2$ is associated with the $x$-term, the major axis of the ellipse is horizontal. This tells us the foci will lie on a horizontal line passing through the center.

2.2 Calculate the Eccentricity of Ellipse $E$

• WHY: The eccentricity is a crucial characteristic that determines the "ovalness" of the ellipse, and it is directly required to find the foci. Furthermore, the problem states the hyperbola's eccentricity is the reciprocal of the ellipse's eccentricity, making $e_E$ a key value.
• Formula: The eccentricity of an ellipse is given by $e_E = \sqrt{1 - \frac{b_E^2}{a_E^2}}$.
• Step-by-step working:
  • Substitute the values $a_E^2 = 100$ and $b_E^2 = 75$ into the formula: $e_E = \sqrt{1 - \frac{75}{100}}$
  • Simplify the fraction: $e_E = \sqrt{1 - \frac{3}{4}}$
  • Perform the subtraction: $e_E = \sqrt{\frac{1}{4}}$
  • Calculate the square root: $e_E = \frac{1}{2}$
  • So, the eccentricity of the ellipse $E$ is $e_E = \frac{1}{2}$.

2.3 Determine the Foci of Ellipse $E$

• WHY: The problem explicitly states that the foci of the hyperbola $H$ coincide with the foci of the ellipse $E$. Finding these points is therefore a critical step, as they will define the hyperbola's center and focal distance.
• Formula: The distance from the center to each focus is $c_E = a_E e_E$. For a horizontal major axis, the foci are located at $(h_E \pm c_E, k_E)$.
• Step-by-step working:
  • First, calculate $c_E$: $c_E = a_E e_E = 10 \times \frac{1}{2} = 5$
  • Now, use the center $(h_E, k_E) = (1,1)$ and $c_E = 5$ to find the foci: $\text{Foci of } E = (1 \pm 5, 1)$
  • Therefore, the foci of the ellipse $E$ are $(6,1)$ and $(-4,1)$.

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3. Analyzing the Hyperbola $H$

Now we use the information derived from the ellipse and the conditions given for the hyperbola $H$.

3.1 Determine the Center and Foci Distance $c_H$ for Hyperbola $H$

• WHY: The problem states that the foci of hyperbola $H$ coincide with those of ellipse $E$. This directly provides the foci of $H$ and, consequently, its center (which is the midpoint of the foci) and the distance from the center to its foci ($c_H$).
• Concept: Foci of $H$ are the same as foci of $E$. The center of a hyperbola is the midpoint of its foci. The distance from the center to a focus is $c_H$.
• Step-by-step working:
  • The foci of hyperbola $H$ are also $(6,1)$ and $(-4,1)$.
  • The center of the hyperbola $(h_H, k_H)$ is the midpoint of these foci: $(h_H, k_H) = \left(\frac{6 + (-4)}{2}, \frac{1 + 1}{2}\right) = \left(\frac{2}{2}, \frac{2}{2}\right) = (1,1)$
  • The distance from the center $(1,1)$ to either focus, say $(6,1)$, gives $c_H$: $c_H = \sqrt{(6-1)^2 + (1-1)^2} = \sqrt{5^2 + 0^2} = 5$
  • Tip: Since the foci lie on a horizontal line ($y=1$), the transverse axis of the hyperbola is also horizontal. This confirms our choice of standard hyperbola equation (with $x$-term positive).

3.2 Calculate Eccentricity $e_H$ of Hyperbola $H$

• WHY: The problem states a direct relationship between the eccentricities of the ellipse and hyperbola, $e_H = 1/e_E$. This allows us to find $e_H$ using the value of $e_E$ we already calculated.
• Given Condition: The eccentricity of the hyperbola $H$ is the reciprocal of the eccentricity of the ellipse $E$.
• Step-by-step working:
  • We found $e_E = \frac{1}{2}$.
  • Therefore, $e_H = \frac{1}{e_E} = \frac{1}{1/2} = 2$.
  • So, the eccentricity of the hyperbola $H$ is $e_H = 2$.

3.3 Determine Semi-Transverse Axis $a_H$ of Hyperbola $H$

• WHY: The length of the transverse axis ($\alpha = 2a_H$) is one of the quantities we need for the final calculation. We can find $a_H$ using the relationship between $c_H$ and $e_H$.
• Formula: For a hyperbola, the distance from the center to a focus is $c_H = a_H e_H$.
• Step-by-step working:
  • We know $c_H = 5$ and $e_H = 2$.
  • Substitute these values into the formula: $5 = a_H \times 2$
  • Solve for $a_H$: $a_H = \frac{5}{2}$
  • Thus, the semi-transverse axis length is $a_H = \frac{5}{2}$.

3.4 Determine Semi-Conjugate Axis $b_H$ of Hyperbola $H$

• WHY: The length of the conjugate axis ($\beta = 2b_H$) is the other quantity we need for the final calculation. We can find $b_H$ using the fundamental relationship between $a_H, b_H$, and $c_H$ for a hyperbola.
• Formula: For a hyperbola, the relationship