1. Interpret the directrix

The given directrix is $5x=4\sqrt{5}\quad \Rightarrow \quad x=\frac{4}{\sqrt{5}}.$

Since the hyperbola is centred at the origin and a directrix is of the form $x=\pm \frac{a}{e}$, the transverse axis is along the $x$-axis.

So the hyperbola must be of the form $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1,$ with directrices $x=\pm \frac{a}{e}.$

Hence, $\frac{a}{e}=\frac{4}{\sqrt{5}} \implies a=\frac{4e}{\sqrt{5}}.$

Therefore, $a^2=\frac{16e^2}{5}.$

2. Use the fact that the hyperbola passes through $(4,-2\sqrt{3})$

Substitute $x=4$, $y=-2\sqrt{3}$ into $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1.$

We get $\frac{16}{a^2}-\frac{12}{b^2}=1.$

For a hyperbola, $e^2=1+\frac{b^2}{a^2} \implies b^2=a^2(e^2-1).$

Substitute this into the point condition: $\frac{16}{a^2}-\frac{12}{a^2(e^2-1)}=1.$

Now use $a^2=\frac{16e^2}{5}$: $\frac{16}{16e^2/5}-\frac{12}{(16e^2/5)(e^2-1)}=1.$

Simplify: $\frac{5}{e^2}-\frac{15}{4e^2(e^2-1)}=1.$

Multiply throughout by $4e^2(e^2-1)$: $20(e^2-1)-15=4e^2(e^2-1).$

So, $20e^2-20-15=4e^4-4e^2,$ $20e^2-35=4e^4-4e^2.$

Bring all terms to one side: $4e^4-24e^2+35=0.$

3. Match with the options

We obtained $4e^4-24e^2+35=0,$ which is Option B.

4. Verification with stored answer

Stored correct answer: B

Derived answer: B

So they agree.