Key Concepts and Formulas

For a hyperbola centered at the origin, we consider two standard forms based on the orientation of its transverse axis.

1. Standard Equation of Hyperbola (Transverse axis along x-axis): If the transverse axis lies along the x-axis, the equation of the hyperbola is given by: $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$
   • Here, $a$ represents the length of the semi-transverse axis and $b$ represents the length of the semi-conjugate axis.
2. Relationship between $a, b,$ and Eccentricity $e$: For a hyperbola, the eccentricity $e$ is always greater than 1 ($e > 1$). The relationship between $a, b,$ and $e$ is: $b^2 = a^2(e^2 - 1)$
3. Equations of Directrices: For a hyperbola with its transverse axis along the x-axis, the equations of its directrices are: $x = \pm \frac{a}{e}$ The directrices are lines perpendicular to the transverse axis.

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Step-by-Step Derivation

Step 1: Identify the orientation of the hyperbola and its directrix equation. We are given that a directrix of the hyperbola is $5x = 4\sqrt{5}$. To make it match the standard form $x = \pm \frac{a}{e}$, we rewrite the given equation: $5x = 4\sqrt{5} \implies x = \frac{4\sqrt{5}}{5}$ Explanation: The form of the directrix ($x = \text{constant}$) immediately tells us that the transverse axis of the hyperbola lies along the x-axis. This is crucial for selecting the correct standard equation and directrix formula.

Step 2: Establish a relationship between $a$ and $e$. Comparing the given directrix equation $x = \frac{4\sqrt{5}}{5}$ with the standard directrix equation $x = \frac{a}{e}$ (we can choose the positive directrix without loss of generality, as 'a' is a length and positive): $\frac{a}{e} = \frac{4\sqrt{5}}{5}$ From this, we can express $a$ in terms of $e$: $a = \frac{4\sqrt{5}}{5}e$ Explanation: This step uses the definition of the directrix to create a fundamental relationship between the semi-transverse axis length ($a$) and the eccentricity ($e$). This relationship will allow us to eliminate $a$ later.

Step 3: Utilize the point the hyperbola passes through. The hyperbola passes through the point $(4, -2\sqrt{3})$. Since this point lies on the hyperbola, it must satisfy its standard equation $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$. Substitute $x=4$ and $y=-2\sqrt{3}$ into the equation: $\frac{(4)^2}{a^2} - \frac{(-2\sqrt{3})^2}{b^2} = 1$ $\frac{16}{a^2} - \frac{12}{b^2} = 1$ Explanation: Any point lying on a curve must satisfy its equation. This provides a second equation involving $a$ and $b$, which we will combine with our previous findings.

Step 4: Substitute $b^2$ in terms of $a$ and $e$. We know the relationship $b^2 = a^2(e^2 - 1)$ for a hyperbola. Substitute this into the equation from Step 3: $\frac{16}{a^2} - \frac{12}{a^2(e^2 - 1)} = 1$ Explanation: This step eliminates $b^2$ from the equation, leaving us with an equation containing only $a$ and $e$. This is a crucial step towards finding an equation solely in terms of $e$.

Step 5: Substitute $a$ in terms of $e$ and simplify. From Step 2, we have $a = \frac{4\sqrt{5}}{5}e$. Let's find $a^2$: $a^2 = \left(\frac{4\sqrt{5}}{5}e\right)^2 = \frac{16 \cdot 5}{25}e^2 = \frac{16}{5}e^2$ Now substitute this value of $a^2$ into the equation from Step 4: $\frac{16}{\frac{16}{5}e^2} - \frac{12}{\frac{16}{5}e^2(e^2 - 1)} = 1$ Simplify the terms: $\frac{16 \cdot 5}{16e^2} - \frac{12 \cdot 5}{16e^2(e^2 - 1)} = 1$ $\frac{5}{e^2} - \frac{60}{16e^2(e^2 - 1)} = 1$ $\frac{5}{e^2} - \frac{15}{4e^2(e^2 - 1)} = 1$ Explanation: This is the core algebraic substitution. By replacing $a^2$ with its expression in terms of $e^2$, we convert the entire equation into a single variable, $e$. Careful algebraic simplification is essential here to avoid errors.

Step 6: Solve the equation for $e$. To eliminate the denominators, find a common denominator, which is $4e^2(e^2 - 1)$. $\frac{5 \cdot 4(e^2 - 1) - 15}{4e^2(e^2 - 1)} = 1$ Multiply both sides by $4e^2(e^2 - 1)$: $20(e^2 - 1) - 15 = 4e^2(e^2 - 1)$ Expand both sides: $20e^2 - 20 - 15 = 4e^4 - 4e^2$ Combine constant terms and rearrange to form a polynomial equation: $20e^2 - 35 = 4e^4 - 4e^2$ Move all terms to one side to set the equation to zero: $4e^4 - 4e^2 - 20e^2 + 35 = 0$ $4e^4 - 24e^2 + 35 = 0$ Explanation: This final step involves standard algebraic manipulation to solve for $e$. The goal is to express the relationship as a polynomial in $e$, which can then be matched with the given options.

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Final Check and Conclusion

The derived equation is $4e^4 - 24e^2 + 35 = 0$. Comparing this with the given options: (A) $4e^4 – 24e^2 + 27 = 0$ (B) $4e^4 – 24e^2 + 35 = 0$ (C) $4e^4 – 12e^2 - 27 = 0$ (D) $4e^4 + 8e^2 - 35 = 0$

Our derived equation matches option (B).

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Tips and Common Mistakes to Avoid

• Orientation is Key: Always determine the orientation of the transverse axis (x-axis or y-axis) first, based on the directrix or foci. This dictates which standard equations to use.
• Hyperbola vs. Ellipse: Remember the relation for eccentricity is $b^2 = a^2(e^2 - 1)$ for a hyperbola ($e > 1$), not $b^2 = a^2(1 - e^2)$ which is for an ellipse ($e < 1$).
• Algebraic Precision: Be extremely careful with squaring terms, handling fractions, and signs during algebraic manipulations. A small error can lead to an incorrect final equation.
• Simplification: Simplify expressions at each stage to make the calculations manageable.

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Summary and Key Takeaway

This problem effectively tests the understanding of fundamental properties of a hyperbola: its standard equation, the definition of its directrix, and the relationship between its parameters ($a, b, e$). The strategy involved using the directrix equation to establish a link between $a$ and $e$, substituting this into the hyperbola's equation (after plugging in the given point), and then using the $b^2 = a^2(e^2-1)$ relation to eliminate $b$. This systematic elimination of variables allowed us to derive a polynomial equation solely in terms of the eccentricity $e$.