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Step-by-Step Solution
Step 1: Identify the Standard Form of the Hyperbola
A hyperbola centered at the origin with its transverse axis along the x-axis can be written as
$ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 $.
Its eccentricity is defined by
$ e^2 = 1 + \frac{b^2}{a^2} $.
Step 2: Use the Given Directrix to Relate $a$ and $e$
For the hyperbola $ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 $, one directrix is given by
$ x = \frac{a}{e} $.
According to the problem, the directrix is
$ 5x = 4\sqrt{5} \;\Longrightarrow\; x = \frac{4\sqrt{5}}{5}. $
Hence,
$ \frac{a}{e} = \frac{4\sqrt{5}}{5} \;\Longrightarrow\; a = e \times \frac{4\sqrt{5}}{5}. $
Step 3: Express $a^2$ and $b^2$ in Terms of $e$
From the above relation:
\[
a = e \cdot \frac{4\sqrt{5}}{5}
\quad\Longrightarrow\quad
a^2 = e^2 \left(\frac{4\sqrt{5}}{5}\right)^2
= e^2 \cdot \frac{16 \times 5}{25}
= \frac{16 e^2}{5}.
\]
Because
$ b^2 = a^2(e^2 - 1) $
for a hyperbola (since $ e^2 = 1 + \tfrac{b^2}{a^2} $), we get
\[
b^2 = \left(\frac{16 e^2}{5}\right)\bigl(e^2 - 1\bigr).
\]
Step 4: Use the Coordinates of the Given Point on the Hyperbola
The hyperbola passes through the point $(4, -2\sqrt{3}).$ Substituting $x=4$ and $y=-2\sqrt{3}$ into
$ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 $
gives
\[
\frac{16}{a^2} - \frac{(2\sqrt{3})^2}{b^2} = 1
\quad\Longrightarrow\quad
\frac{16}{a^2} - \frac{12}{b^2} = 1.
\]
Step 5: Substitute $a^2$ and $b^2$ into the Above Equation
Using
$ a^2 = \frac{16 e^2}{5} $
and
$ b^2 = \frac{16 e^2}{5}(e^2 - 1) $,
we get
\[
\frac{16}{\frac{16 e^2}{5}} \;-\; \frac{12}{\frac{16 e^2}{5}(\,e^2 - 1)} = 1.
\]
Simplify each term:
1)
$\displaystyle \frac{16}{\frac{16 e^2}{5}}
= \frac{16 \times 5}{16 e^2}
= \frac{5}{e^2}.$
2)
$\displaystyle \frac{12}{\frac{16 e^2}{5}(e^2 - 1)}
= \frac{12 \times 5}{16 e^2 (e^2 - 1)}
= \frac{60}{16 e^2 (e^2 - 1)}
= \frac{15}{4 e^2 (e^2 - 1)}.$
So the equation becomes
\[
\frac{5}{e^2} \;-\; \frac{15}{4 e^2 (e^2 - 1)} = 1.
\]
Step 6: Clear the Denominator to Form an Equation in $e$
Multiply through by $4 e^2(e^2 - 1)$:
\[
4 e^2(e^2 - 1)\,\Bigl(\frac{5}{e^2}\Bigr)
\;-\;
4 e^2(e^2 - 1)\,\Bigl(\frac{15}{4 e^2 (e^2 - 1)}\Bigr)
=
4 e^2(e^2 - 1)\,(1).
\]
Simplify each part separately:
• First product:
\[
4 e^2(e^2 - 1)\,\Bigl(\frac{5}{e^2}\Bigr)
= 5 \times 4 (e^2 - 1)
= 20(e^2 - 1)
= 20 e^2 - 20.
\]
• Second product:
\[
4 e^2(e^2 - 1)\,\Bigl(\frac{15}{4 e^2 (e^2 - 1)}\Bigr)
= 15.
\]
Hence the left side is:
\[
\bigl(20 e^2 - 20\bigr) - 15
= 20 e^2 - 35.
\]
The right side is:
\[
4 e^2(e^2 - 1)
= 4 e^4 - 4 e^2.
\]
So the equation becomes
\[
20 e^2 - 35 = 4 e^4 - 4 e^2.
\]
Rearrange to set everything to zero:
\[
4 e^4 - 4 e^2 - 20 e^2 + 35 = 0
\;\Longrightarrow\;
4 e^4 - 24 e^2 + 35 = 0.
\]
Step 7: Conclusion
We have arrived at the required equation for the eccentricity:
\[
\boxed{4 e^4 - 24 e^2 + 35 = 0}.
\]
Thus, the correct option is
$4e^4 - 24e^2 + 35 = 0.$
