This problem requires a thorough understanding of the fundamental properties of a hyperbola, specifically its eccentricity, the length of its latus rectum, and the condition for a line to be tangent to it. We will systematically use the given information to first determine the defining parameters of the hyperbola ($a^2$ and $b^2$), and then apply the tangency condition to find the required value of $c^2$.

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1. Understanding the Problem and Identifying Key Information

We are given a hyperbola in its standard form: $H: \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ Our goal is to find the value of $c^2$ for a line $y = 2x + c$ that is tangent to this hyperbola. We are provided with the following characteristics of the hyperbola:

• Its eccentricity ($e$) is $\sqrt{\frac{5}{2}}$.
• The length of its latus rectum (LLR) is $6\sqrt{2}$.
• The tangent line is $y = 2x + c$.

The strategy will be to use the first two pieces of information (eccentricity and LLR) to determine the values of $a^2$ and $b^2$, and then use these values along with the tangent line equation in the tangency condition formula to find $c^2$.

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2. Utilizing Eccentricity to Establish a Relationship between $a^2$ and $b^2$

Key Concept: For a hyperbola with the standard equation $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, its eccentricity ($e$) is related to the semi-transverse axis ($a$) and semi-conjugate axis ($b$) by the formula: $e^2 = 1 + \frac{b^2}{a^2}$

Why this step? The eccentricity is a fundamental property that defines the shape of the hyperbola. This formula directly connects $e$ with $a$ and $b$, providing our first algebraic equation relating these two unknown parameters.

Step-by-step working:

1. We are given the eccentricity $e = \sqrt{\frac{5}{2}}$.
2. Substitute this value into the eccentricity formula: $\left(\sqrt{\frac{5}{2}}\right)^2 = 1 + \frac{b^2}{a^2}$
3. Simplify the equation: $\frac{5}{2} = 1 + \frac{b^2}{a^2}$
4. Isolate the term $\frac{b^2}{a^2}$: $\frac{b^2}{a^2} = \frac{5}{2} - 1$ $\frac{b^2}{a^2} = \frac{5 - 2}{2}$ $\frac{b^2}{a^2} = \frac{3}{2}$
5. From this, we get our first relationship between $a^2$ and $b^2$: $b^2 = \frac{3}{2}a^2 \quad \text{(Equation 1)}$

Tip for JEE: Always remember that for a hyperbola, the eccentricity $e$ must be greater than 1 ($e > 1$). If your calculation yields $e \le 1$, it indicates an error or that the curve is an ellipse/circle, not a hyperbola.

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3. Using the Length of Latus Rectum to Determine $a^2$ and $b^2$

Key Concept: The length of the latus rectum (LLR) for a hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is given by the formula: $\text{LLR} = \frac{2b^2}{a}$

Why this step? The LLR provides a second independent equation relating $a$ and $b$. With two equations (Equation 1 from eccentricity and this new equation from LLR) and two unknowns ($a^2$ and $b^2$), we can now solve for their specific values.

Step-by-step working:

1. We are given that the length of the latus rectum is $6\sqrt{2}$.
2. Substitute this value into the LLR formula: $6\sqrt{2} = \frac{2b^2}{a}$
3. Now, substitute the expression for $b^2$ from Equation 1 ($b^2 = \frac{3}{2}a^2$) into this equation: $6\sqrt{2} = \frac{2 \left(\frac{3}{2}a^2\right)}{a}$
4. Simplify the expression: $6\sqrt{2} = \frac{3a^2}{a}$ $6\sqrt{2} = 3a$
5. Solve for $a$: $a = \frac{6\sqrt{2}}{3}$ $a = 2\sqrt{2}$
6. Now, calculate $a^2$: $a^2 = (2\sqrt{2})^2 = 4 \times 2 = 8$
7. Finally, substitute the value of $a^2$ back into Equation 1 to find $b^2$: $b^2 = \frac{3}{2}a^2 = \frac{3}{2}(8)$ $b^2 = 3 \times 4 = 12$

So, we have found the parameters of the hyperbola: $a^2 = 8$ and $b^2 = 12$.

Common Mistake: Be careful not to confuse $a$ with $a^2$ in the LLR formula. The denominator is $a$, not $a^2$. Also, always remember that $a$ represents a length, so it must be positive.

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4. Applying the Tangency Condition to Find $c^2$

Key Concept: For a line $y = mx + c$ to be tangent to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, the following condition must be satisfied: $c^2 = a^2m^2 - b^2$

Why this step? This formula directly relates the slope ($m$) and y-intercept ($c$) of the tangent line to the hyperbola's parameters ($a^2$ and $b^2$). We have already calculated $a^2$ and $b^2$, and the slope $m$ is given in the tangent line equation. This allows us to directly calculate $c^2$.

Step-by-step working:

1. The given tangent line is $y = 2x + c$.
2. From this equation, we can identify the slope $m = 2$.
3. We have previously calculated $a^2 = 8$ and $b^2 = 12$.
4. Substitute these values into the tangency condition formula: $c^2 = (8)(2^2) - 12$ $c^2 = (8)(4) - 12$ $c^2 = 32 - 12$ $c^2 = 20$

Important Note: It's crucial to remember the correct sign in the tangency condition. For a hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, it is $c^2 = a^2m^2 - b^2$. For an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, the condition is $c^2 = a^2m^2 + b^2$. A common error is mixing these two up.

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5. Conclusion and Final Answer

Based on our calculations, the value of $c^2$ is 20.

Comparing this with the given options: (A) 18 (B) 20 (C) 24 (D) 32

The correct option is (B).

Key Takeaway: This problem demonstrates a typical approach to solving coordinate geometry questions involving conic sections. It involves:

1. Extracting information from the problem statement (eccentricity, LLR, tangent line).
2. Applying relevant formulas for the specific conic section (hyperbola in this case) to establish relationships between its parameters.
3. Solving a system of equations to find the unknown parameters ($a^2, b^2$).
4. Using a final condition (tangency) with the determined parameters to find the required value. Always ensure you use the correct formulas and pay attention to signs, especially when dealing with different conic sections.