This problem is a comprehensive test of your understanding of hyperbolas, requiring you to utilize their fundamental properties, the general equation of a tangent, and coordinate geometry concepts. We will systematically construct the hyperbola's equation, derive the specific tangent line based on given conditions, calculate its intercepts, and finally evaluate the required expression.

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1. Establishing the Hyperbola's Standard Equation

The journey begins by determining the specific equation of the hyperbola using the provided foci and eccentricity.

Key Concept: The standard equation of a hyperbola centered at the origin with its foci on the x-axis is given by: $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ where $a$ is the length of the semi-transverse axis and $b$ is the length of the semi-conjugate axis. For such a hyperbola, the foci are located at $(\pm ae, 0)$, and the eccentricity $e$ is related to $a$ and $b$ by the equation: $b^2 = a^2(e^2 - 1)$

Step 1.1: Relate Given Foci and Eccentricity to Hyperbola Parameters.

• Given Foci: The foci are provided as $(\pm 2, 0)$.
  • Why: By comparing this with the standard form of foci $(\pm ae, 0)$, we can directly establish the relation: $ae = 2$
• Given Eccentricity: The eccentricity is given as $e = \frac{3}{2}$.

Step 1.2: Calculate the Semi-Transverse Axis Length, 'a'.

• Action: Substitute the given value of $e$ into the relation $ae = 2$: $a \left(\frac{3}{2}\right) = 2$
• Calculation: Solve for $a$: $a = \frac{2 \times 2}{3} \implies a = \frac{4}{3}$
  • Why: Determining 'a' is a fundamental step as it defines the extent of the hyperbola along its transverse axis and is crucial for finding $b^2$.
  • Tip: Be extremely careful not to confuse this 'a' (representing the semi-transverse axis of the hyperbola) with the 'a' that denotes the x-intercept in the problem statement later on. They are distinct variables in this problem context.

Step 1.3: Calculate the Square of the Semi-Conjugate Axis Length, 'b²'.

• Action: Use the fundamental relationship between $a$, $b$, and $e$ for a hyperbola: $b^2 = a^2(e^2 - 1)$
  • Why: This formula is essential for defining the shape and dimensions of the hyperbola, allowing us to complete its equation.
  • Common Mistake: For a hyperbola, the term inside the parenthesis is $(e^2 - 1)$. Remember that for an ellipse, the formula is $b^2 = a^2(1 - e^2)$. Using the wrong formula is a common error.
• Calculation: Substitute the calculated value $a = \frac{4}{3}$ (which means $a^2 = \frac{16}{9}$) and the given $e = \frac{3}{2}$ (which means $e^2 = \frac{9}{4}$): $b^2 = \left(\frac{4}{3}\right)^2 \left(\left(\frac{3}{2}\right)^2 - 1\right)$ $b^2 = \frac{16}{9} \left(\frac{9}{4} - 1\right)$ $b^2 = \frac{16}{9} \left(\frac{9-4}{4}\right)$ $b^2 = \frac{16}{9} \left(\frac{5}{4}\right)$ $b^2 = \frac{4 \times 5}{9} \implies b^2 = \frac{20}{9}$

Step 1.4: Write the Equation of the Hyperbola.

• Action: Substitute the calculated values $a^2 = \frac{16}{9}$ and $b^2 = \frac{20}{9}$ into the standard equation $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$: $\frac{x^2}{16/9} - \frac{y^2}{20/9} = 1$ $\frac{9x^2}{16} - \frac{9y^2}{20} = 1$
  • Why: This is the specific algebraic representation of the hyperbola we are dealing with throughout the problem.

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2. Determining the Equation of the Tangent Line

Next, we need to find the equation of a tangent line that satisfies the given conditions: it's perpendicular to a specific line and touches the hyperbola in the first quadrant.

Key Concept: The equation of a tangent with slope $m$ to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is given by: $y = mx \pm \sqrt{a^2m^2 - b^2}$

Step 2.1: Find the Slope of the Tangent Line.

• Given Information: The tangent is perpendicular to the line $2x + 3y = 6$.
• Action: First, find the slope of the given line. Rewrite its equation in the slope-intercept form $y = m_1x + c_1$: $3y = -2x + 6$ $y = -\frac{2}{3}x + 2$ The slope of this line is $m_1 = -\frac{2}{3}$.
• Calculation: If two lines are perpendicular, the product of their slopes is $-1$. Let the slope of the tangent be $m$. $m \cdot m_1 = -1$ $m \left(-\frac{2}{3}\right) = -1$ $m = \frac{3}{2}$
  • Why: The slope is a crucial parameter needed to use the general tangent equation formula.

Step 2.2: Apply the General Tangent Equation Formula.

• Action: Substitute the known values: $a^2 = \frac{16}{9}$, $b^2 = \frac{20}{9}$, and the calculated slope $m = \frac{3}{2}$ into the general tangent formula: $y = mx \pm \sqrt{a^2m^2 - b^2}$
  • Why: This formula directly provides the equation(s) of lines that are tangent to the hyperbola with the specific slope $m$.
  • Common Mistake: Again, ensure you use the correct formula for a hyperbola. For an ellipse, the term under the square root would be $a^2m^2 + b^2$.
• Calculation: $y = \frac{3}{2}x \pm \sqrt{\left(\frac{16}{9}\right)\left(\frac{3}{2}\right)^2 - \frac{20}{9}}$ $y = \frac{3}{2}x \pm \sqrt{\frac{16}{9} \cdot \frac{9}{4} - \frac{20}{9}}$ $y = \frac{3}{2}x \pm \sqrt{4 - \frac{20}{9}}$ $y = \frac{3}{2}x \pm \sqrt{\frac{36 - 20}{9}}$ $y = \frac{3}{2}x \pm \sqrt{\frac{16}{9}}$ $$$$