Key Concepts and Formulae for Hyperbola

To solve this problem effectively, we need a strong understanding of the standard form of a hyperbola and how its eccentricity is defined.

1. Standard Equation of a Hyperbola (Center at Origin, Transverse Axis along X-axis): A hyperbola with its center at the origin $(0,0)$ and its transverse axis lying along the x-axis (meaning its vertices are on the x-axis) has the standard equation: $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ Here:

   • $a$ represents the length of the semi-transverse axis. The vertices of the hyperbola are located at $(\pm a, 0)$.
   • $b$ represents the length of the semi-conjugate axis.
   • The total length of the transverse axis is $2a$.
   • The total length of the conjugate axis is $2b$.

2. Eccentricity of a Hyperbola: For a hyperbola of the form $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, the eccentricity, denoted by $e$, is a fundamental characteristic that describes the "openness" or "flatness" of the hyperbola's branches. It is always greater than 1 ($e > 1$). The formula relating $a$, $b$, and $e$ is: $b^2 = a^2(e^2 - 1)$ From this, we can derive the formula for eccentricity: $e^2 - 1 = \frac{b^2}{a^2}$ $e^2 = 1 + \frac{b^2}{a^2}$ $e = \sqrt{1 + \frac{b^2}{a^2}}$ This formula will be crucial for our final calculation.

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Step-by-Step Solution

1. Identify the Standard Equation of the Hyperbola

• Why this step? The first logical step in solving any problem involving conic sections is to establish the correct algebraic representation (standard equation) based on the given geometric properties. This provides the framework for using all other information.
• The problem explicitly states that the hyperbola has its center at the origin $(0,0)$ and its transverse axis lies along the x-axis.
• Based on our key concepts, the standard equation for such a hyperbola is: $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ Here, $a$ is associated with the x-term, indicating the transverse axis is horizontal.

2. Determine the Value of the Semi-Transverse Axis 'a'

• Why this step? The length of the transverse axis is a direct measure of $2a$. Finding $a$ (and subsequently $a^2$) is essential because $a^2$ is a parameter in the hyperbola's equation and also directly used in the eccentricity formula.
• The problem provides that the length of the transverse axis is 4.
• By definition, the length of the transverse axis is $2a$.
• Therefore, we can set up the equation: $2a = 4$
• Dividing both sides by 2, we find the value of $a$: $a = 2$
• To use this in the standard equation and eccentricity formula, we need $a^2$: $a^2 = 2^2 = 4$

3. Determine the Value of the Semi-Conjugate Axis 'b' (or $b^2$)

• Why this step? At this point, we have one unknown parameter left in the hyperbola's equation: $b^2$. The problem states that the hyperbola passes through the point $(4, 2)$. If a point lies on the hyperbola, its coordinates must satisfy the hyperbola's equation. By substituting the coordinates of this point and the $a^2$ value we just found, we can solve for $b^2$.
• The hyperbola passes through the point $(4, 2)$. This means we can substitute $x=4$ and $y=2$ into our standard equation, along with the value $a^2=4$: $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ $\frac{4^2}{4} - \frac{2^2}{b^2} = 1$
• Now, let's simplify and solve for $b^2$: $\frac{16}{4} - \frac{4}{b^2} = 1$ $4 - \frac{4}{b^2} = 1$
• Subtract 4 from both sides of the equation: $- \frac{4}{b^2} = 1 - 4$ $- \frac{4}{b^2} = -3$
• Multiply both sides by -1 to make the terms positive: $\frac{4}{b^2} = 3$
• To isolate $b^2$, we can cross-multiply or swap $b^2$ with 3: $b^2 = \frac{4}{3}$

4. Calculate the Eccentricity 'e'

• Why this step? This is the final step to answer the question posed. We have successfully determined the values of $a^2$ and $b^2$, which are the two parameters required to calculate the eccentricity using its defining formula.
• The formula for the eccentricity $e$ of this type of hyperbola is: $e = \sqrt{1 + \frac{b^2}{a^2}}$
• Now, substitute the values we found: $a^2 = 4$ and $b^2 = \frac{4}{3}$ into the formula: $e = \sqrt{1 + \frac{4/3}{4}}$
• Simplify the complex fraction $\frac{4/3}{4}$. Remember that dividing by a number is equivalent to multiplying by its reciprocal: $\frac{4/3}{4} = \frac{4}{3} \times \frac{1}{4} = \frac{4}{12} = \frac{1}{3}$
• Substitute this simplified fraction back into the eccentricity formula: $e = \sqrt{1 + \frac{1}{3}}$
• Combine the terms under the square root by finding a common denominator: $e = \sqrt{\frac{3}{3} + \frac{1}{3}}$ $e = \sqrt{\frac{4}{3}}$
• Finally, take the square root of the numerator and the denominator: $e = \frac{\sqrt{4}}{\sqrt{3}}$ $e = \frac{2}{\sqrt{3}}$

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Relevant Tips and Common Mistakes

• Transverse Axis Orientation: Always double-check whether the transverse axis is along the x-axis or y-axis. If it were along the y-axis, the standard equation would be $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$. The parameters $a$ and $b$ always refer to the semi-transverse and semi-conjugate axes respectively, but their positions under $x^2$ and $y^2$ change.
• Eccentricity Formula for Hyperbola vs. Ellipse: This is a very common source of error. Be extremely careful not to confuse the formulas:
  • For a hyperbola: $e = \sqrt{1 + \frac{b^2}{a^2}}$. Note that $e > 1$.
  • For an ellipse: $e = \sqrt{1 - \frac{b^2}{a^2}}$ (where $a$ is the semi-major axis, and $a>b$). Note that $0 \le e < 1$. The plus sign for hyperbola is a key differentiator.
• Algebraic Precision: Pay close attention to arithmetic, especially when simplifying fractions (like $\frac{4/3}{4}$) and handling negative signs. A small calculation error can lead to an incorrect final answer.
• Understanding 'a' and 'b' (Hyperbola Specific): For a hyperbola, $a$ is always the semi-length of the transverse axis (the one associated with the positive term in the standard equation), and $b$ is always the semi-length of the conjugate axis.

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Summary and Key Takeaway

This problem is a straightforward application of the standard form and eccentricity definition for a hyperbola. The methodical approach involves:

1. Identifying the correct standard equation based on the given center and transverse axis orientation.
2. Using the length of the transverse axis to directly determine the value of $a$.
3. Utilizing the given point through which the hyperbola passes to solve for $b$ (or $b^2$) by substituting the point's coordinates into the equation.
4. Finally, applying the eccentricity formula with the determined $a^2$ and $b^2$ values to calculate $e$.

The final answer is $\boxed{\frac{2}{\sqrt{3}}}$.