Here is a comprehensive and elaborate solution, designed to be educational and clear for JEE aspirants.

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This problem requires a deep understanding of the fundamental properties of ellipses and hyperbolas, including their foci, eccentricity, and latus rectum. We will systematically use the given information to determine the parameters of both conic sections and then calculate the required sum.

1. Understanding the Conics and Their Key Formulas

Let's first establish the standard forms and essential properties for an ellipse and a hyperbola centered at the origin, with their major/transverse axis along the x-axis, as implied by the problem notation.

a) Ellipse (E): $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, where $a>b>0$.

• Semi-major axis: $a$
• Semi-minor axis: $b$
• Foci: The foci are located at $(\pm ae_E, 0)$, where $e_E$ is the eccentricity of the ellipse.
• Distance between foci: $2ae_E$.
• Relationship between $a, b, e_E$: The semi-minor axis $b$ is related to $a$ and $e_E$ by $b^2 = a^2(1-e_E^2)$. This is often rearranged as $a^2e_E^2 = a^2-b^2$.
• Eccentricity: $e_E = \frac{\sqrt{a^2-b^2}}{a}$. For an ellipse, $0 < e_E < 1$.
• Length of Latus Rectum: $L_E = \frac{2b^2}{a}$.

b) Hyperbola (H): $\frac{x^2}{A^2}-\frac{y^2}{B^2}=1$.

• Semi-transverse axis: $A$
• Semi-conjugate axis: $B$
• Foci: The foci are located at $(\pm Ae_H, 0)$, where $e_H$ is the eccentricity of the hyperbola.
• Distance between foci: $2Ae_H$.
• Relationship between $A, B, e_H$: The semi-conjugate axis $B$ is related to $A$ and $e_H$ by $B^2 = A^2(e_H^2-1)$. This is often rearranged as $A^2e_H^2 = A^2+B^2$.
• Eccentricity: $e_H = \frac{\sqrt{A^2+B^2}}{A}$. For a hyperbola, $e_H > 1$.
• Length of Latus Rectum: $L_H = \frac{2B^2}{A}$.

2. Setting Up Equations from the Foci Distances

The problem states that "the distance between the foci of E and the foci of H is $2\sqrt{3}$." In standard JEE problems, this phrasing implies that the focal length (distance between foci) for each conic section is $2\sqrt{3}$. If it meant the distance between a focus of E and a focus of H, it would be specified more clearly (e.g., "distance between nearest foci").

a) For the Ellipse E:

• Concept: The distance between the foci of an ellipse is $2ae_E$.
• Action: We equate this to the given value: $2ae_E = 2\sqrt{3}$
• Explanation: This equation directly relates the semi-major axis $a$ and eccentricity $e_E$ of the ellipse to the given focal distance.
• Dividing by 2, we get: $ae_E = \sqrt{3}$
• Action: To bring in the semi-minor axis $b$, we use the fundamental relationship $a^2e_E^2 = a^2-b^2$.
• Explanation: This relationship is derived from the definition of eccentricity for an ellipse ($e_E = \frac{\sqrt{a^2-b^2}}{a}$) and is crucial for connecting $a, b,$ and $e_E$.
• Squaring both sides of $ae_E = \sqrt{3}$ gives $a^2e_E^2 = (\sqrt{3})^2 = 3$.
• Therefore, for the ellipse, we have our first key equation: $a^2-b^2 = 3 \quad \text{(Equation 1)}$

b) For the Hyperbola H:

• Concept: The distance between the foci of a hyperbola is $2Ae_H$.
• Action: We equate this to the given value: $2Ae_H = 2\sqrt{3}$
• Explanation: Similar to the ellipse, this equation relates the semi-transverse axis $A$ and eccentricity $e_H$ of the hyperbola to the given focal distance.
• Dividing by 2, we get: $Ae_H = \sqrt{3}$
• Action: To bring in the semi-conjugate axis $B$, we use the fundamental relationship $A^2e_H^2 = A^2+B^2$.
• Explanation: This relationship is derived from the definition of eccentricity for a hyperbola ($e_H = \frac{\sqrt{A^2+B^2}}{A}$) and is crucial for connecting $A, B,$ and $e_H$.
• Squaring both sides of $Ae_H = \sqrt{3}$ gives $A^2e_H^2 = (\sqrt{3})^2 = 3$.
• Therefore, for the hyperbola, we have our second key equation: $A^2+B^2 = 3 \quad \text{(Equation 2)}$

Tip for avoiding common mistakes: Notice the crucial difference in the relationships: for an ellipse, $a^2e_E^2 = a^2-b^2$ (subtraction), while for a hyperbola, $A^2e_H^2 = A^2+B^2$ (addition). This sign difference is fundamental and arises directly from their geometric definitions and the fact that $e_E < 1$ and $e_H > 1$.

3. Utilizing the Eccentricity Ratio and Semi-Axis Difference

We are given two more pieces of information that will help us determine the values of $a$ and $A$:

1. The difference between the semi-major axis of the ellipse and the semi-transverse axis of the hyperbola: $a-A=2$.
2. The ratio of their eccentricities: $\frac{e_E}{e_H}=\frac{1}{3}$.

a) Expressing Eccentricities in terms of $a$ and $A$:

• Action: From the results of Step 2, we have $ae_E = \sqrt{3}$ and $Ae_H = \sqrt{3}$. We can express the eccentricities as:
  • $e_E = \frac{\sqrt{3}}{a}$
  • $e_H = \frac{\sqrt{3}}{A}$
• Explanation: We isolate $e_E$ and $e_H$ because the next step involves their ratio.

b) Using the Eccentricity Ratio:

• Action: Substitute these expressions for $e_E$ and $e_H$ into the given eccentricity ratio $\frac{e_E}{e_H}=\frac{1}{3}$: $\frac{\frac{\sqrt{3}}{a}}{\frac{\sqrt{3}}{A}} = \frac{1}{3}$
• Explanation: This step allows us to relate $a$ and $A$ using the given ratio.
• Simplifying the complex fraction: $\frac{\sqrt{3}}{a} \times \frac{A}{\sqrt{3}} = \frac{A}{a}$
• So, we have: $\frac{A}{a} = \frac{1}{3}$
• This gives us a direct relationship between $a$ and $A$: $a = 3A \quad \text{(Equation 3)}$

c) Solving for $a$ and $A$:

• Concept: We now have a system of two linear equations involving $a$ and $A$:
  1. $a-A=2$ (Given)
  2. $a=3A$ (From Equation 3)
• Action: Substitute Equation 3 into the first equation: $(3A) - A = 2$ $2A = 2$
• Explanation: This substitution eliminates $a$, allowing us to solve for $A$.
• Solving for $A$: $A = 1$
• Action: Substitute $A=1$ back into Equation 3 to find $a$: $a = 3(1)$ $a = 3$
• Explanation: We have now determined the specific values for the semi-major axis of the ellipse ($a=3$) and the semi-transverse axis of the hyperbola ($A=1$). These values are foundational for calculating the latus rectums.

4. Calculating $b^2$ and $B^2$

With the values of $a$ and $A$ now known, we can use Equation 1 and Equation 2 (derived in Step 2) to find $b^2$ and $B^2$.

a) For the Ellipse E:

• Concept: Use Equation 1: $a^2-b^2=3$.
• Action: Substitute $a=3$: $(3)^2 - b^2 = 3$ $9 - b^2 = 3$
• Explanation: This allows us to find $b^2$, which is needed for the ellipse's latus rectum formula.
• Solving for $b^2$: $b^2 = 9 - 3$ $b^2 = 6$

b) For the Hyperbola H:

• Concept: Use Equation 2: $A^2+B^2=3$.
• Action: Substitute $A=1$: $(1)^2 + B^2 = 3$ $1 + B^2 = 3$
• Explanation: This allows us to find $B^2$, which is needed for the hyperbola's latus rectum formula.
• Solving for $B^2$: