This solution will guide you through the problem step-by-step, explaining the underlying concepts and formulas for ellipses and hyperbolas, which are fundamental to coordinate geometry in JEE Mathematics.

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1. Fundamental Concepts: Ellipse and Hyperbola Properties

To tackle problems involving conic sections, a strong grasp of their standard equations and associated properties like eccentricity, foci, and their distances is essential.

• Ellipse (Major Axis along x-axis):

  • Standard Equation: $\frac{x^2}{A^2} + \frac{y^2}{B^2} = 1$, where $A > B$.
  • Semi-major axis: $a = A$ (along the x-axis).
  • Semi-minor axis: $b = B$ (along the y-axis).
  • Eccentricity ($e$): A measure of the ellipse's "flatness". It's defined by the relation $b^2 = a^2(1 - e^2)$. This implies $e = \sqrt{1 - \frac{b^2}{a^2}}$.
    • Key Property: For an ellipse, $0 < e < 1$.
  • Foci: The two fixed points are located at $(\pm ae, 0)$ on the major axis.
  • Distance between Foci: The distance between $(\text{-}ae, 0)$ and $(ae, 0)$ is $2ae$.

• Hyperbola (Transverse Axis along x-axis):

  • Standard Equation: $\frac{x^2}{A^2} - \frac{y^2}{B^2} = 1$.
  • Semi-transverse axis: $a = A$ (along the x-axis).
  • Semi-conjugate axis: $b = B$ (along the y-axis).
  • Eccentricity ($e$): A measure of how "open" the hyperbola's branches are. It's defined by the relation $b^2 = a^2(e^2 - 1)$. This implies $e = \sqrt{1 + \frac{b^2}{a^2}}$.
    • Key Property: For a hyperbola, $e > 1$.
  • Foci: The two fixed points are located at $(\pm ae, 0)$ on the transverse axis.
  • Distance between Foci: The distance between $(\text{-}ae, 0)$ and $(ae, 0)$ is $2ae$.

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2. Analyzing the Ellipse

Let's consider the given ellipse equation and extract its properties.

Given Ellipse Equation: $\frac{x^2}{25} + \frac{y^2}{b^2} = 1$

Step 1: Identify Standard Parameters ($a_1, b_1$). We compare this with the standard form $\frac{x^2}{A^2} + \frac{y^2}{B^2} = 1$. Here, $A^2 = 25 \implies A = 5$ and $B^2 = b^2 \implies B = b$.

The problem states that $b < 5$. Since $A=5$ and $B=b$ (with $b < 5$), we have $A > B$. This tells us that the major axis of the ellipse lies along the x-axis. Therefore, for the ellipse:

• Semi-major axis, $a_1 = 5$.
• Semi-minor axis, $b_1 = b$. Let $e_1$ be the eccentricity of this ellipse.

Step 2: Express Eccentricity ($e_1$) in terms of $b$. The relationship between the semi-axes and eccentricity for an ellipse is $b_1^2 = a_1^2(1 - e_1^2)$. Substituting the values $a_1=5$ and $b_1=b$: $b^2 = 25(1 - e_1^2) \quad \text{(Equation 1)}$ This equation links the unknown $b^2$ to the eccentricity $e_1$. We know $0 < e_1 < 1$.

Step 3: Define the Distance between Foci ($\alpha$). The distance between the foci of an ellipse with its major axis along the x-axis is $2ae$. For our ellipse, this distance is $\alpha = 2a_1e_1$. $\alpha = 2(5)e_1 = 10e_1 \quad \text{(Equation 2)}$

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3. Analyzing the Hyperbola

Next, let's analyze the given hyperbola equation.

Given Hyperbola Equation: $\frac{x^2}{16} - \frac{y^2}{b^2} = 1$

Step 1: Identify Standard Parameters ($a_2, b_2$). We compare this with the standard form $\frac{x^2}{A^2} - \frac{y^2}{B^2} = 1$. Here, $A^2 = 16 \implies A = 4$ and $B^2 = b^2 \implies B = b$.

For a hyperbola of this form, the transverse axis lies along the x-axis. Therefore, for the hyperbola:

• Semi-transverse axis, $a_2 = 4$.
• Semi-conjugate axis, $b_2 = b$. Let $e_2$ be the eccentricity of this hyperbola.

Step 2: Express Eccentricity ($e_2$) in terms of $b$. The relationship between the semi-axes and eccentricity for a hyperbola is $b_2^2 = a_2^2(e_2^2 - 1)$. Substituting the values $a_2=4$ and $b_2=b$: $b^2 = 16(e_2^2 - 1) \quad \text{(Equation 3)}$ This equation links the same unknown $b^2$ to the eccentricity $e_2$. We know $e_2 > 1$.

Step 3: Define the Distance between Foci ($\beta$). The distance between the foci of a hyperbola with its transverse axis along the x-axis is $2ae$. For our hyperbola, this distance is $\beta = 2a_2e_2$. $\beta = 2(4)e_2 = 8e_2 \quad \text{(Equation 4)}$

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4. Solving for Eccentricities and $b^2$ using the Given Condition

We are provided with a crucial condition: $e_1 e_2 = 1$. This relationship will allow us to find the specific values of $e_1$ and $e_2$, and subsequently $b^2$.

Step 1: Equate the expressions for $b^2$. We have two different expressions for $b^2$ from Equation 1 (ellipse) and Equation 3 (hyperbola). By equating them, we can form an equation involving only $e_1$ and $e_2$: $25(1 - e_1^2) = 16(e_2^2 - 1)$

Step 2: Substitute $e_2 = \frac{1}{e_1}$ into the combined equation. The condition $e_1 e_2 = 1$ implies $e_2 = \frac{1}{e_1}$. Substituting this into the equation from Step 1 will allow us to solve for $e_1$. $25(1 - e_1^2) = 16\left(\left(\frac{1}{e_1}\right)^2 - 1\right)$ $25(1 - e_1^2) = 16\left(\frac{1}{e_1^2} - 1\right)$ $25(1 - e_1^2) = 16\left(\frac{1 - e_1^2}{e_1^2}\right)$

Step 3: Solve for $e_1$. We observe the term $(1 - e_1^2)$ on both sides of the equation. Important Tip: When canceling terms from both sides of an equation, always consider if the term could be zero. If it were zero, canceling it would lead to a loss of a potential solution. For an ellipse, we know $0 < e_1 < 1$. This means $e_1^2 < 1$, so $1 - e_1^2$ is always a positive, non-zero value. Therefore, we can safely cancel $(1 - e_1^2)$ from both sides without losing any valid solutions: $25 = \frac{16}{e_1^2}$ Now, we solve for $e_1^2$: $e_1^2 = \frac{16}{25}$ Taking the square root and remembering that eccentricity is always positive: $e_1 = \sqrt{\frac{16}{25}} = \frac{4}{5}$ This value $e_1 = \frac{4}{5}$ satisfies $0 < e_1 < 1$, which is consistent for an ellipse.

Step 4: Calculate $e_2$. Using the given condition $e_1 e_2 = 1$: $e_2 = \frac{1}{e_1} = \frac{1}{4/5} = \frac{5}{4}$ This value $e_2 = \frac{5}{4}$ satisfies $e_2 > 1$, which is consistent for a hyperbola.

Step 5: Calculate the value of $b^2$. Although not directly asked for $\alpha$ and $\beta$, finding $b^2$ helps verify our calculations and ensures all conditions are met. Substitute $e_1 = \frac{4}{5}$ back into Equation 1: $b^2 = 25(1 - e_1^2) = 25\left(1 - \left(\frac{4}{5}\right)^2\right)$ $b^2 = 25\left(1 - \frac{16}{25}\right) = 25\left(\frac{25 - 16}{25}\right)$ $b^2 = 25\left(\frac{9}{25}\right) = 9$ Thus, $b = \sqrt{9} = 3$. This value $b=3$ satisfies the initial condition $b < 5$.

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5. Calculating the Distances between Foci ($\alpha$ and $\beta$)

Now that we have the eccentricities, we can calculate $\alpha$ and $\beta$.

**Step 1