If a directrix of a hyperbola centered at the origin and passing through the point 4 , - 2 3 is 5 x = 4 5…

Let equation of hyperbola is $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

$\therefore$ It passes through $\left(4, -2\sqrt{3}\right) \Rightarrow \frac{16}{a^2}-\frac{12}{b^2}=1$

$\Rightarrow 16-12\times \frac{a^2}{b^2}=a^2...\left(1\right)$

Equation of directrix is $x=\frac{a}{e}=\frac{4}{\sqrt{5}}$, given in question.

$\Rightarrow a^2=\frac{16}{5}{e}^2...\left(2\right)$

And we know that $b^2=a^2\left(e^2-1\right)$

$\Rightarrow \frac{b^2}{a^2}=e^2-1...\left(3\right)$

$\therefore$ From $\left(1\right),\left(2\right) \& \left(3\right)$

$16-\frac{12}{e^2-1}=\frac{16}{5}{e}^2$

$\Rightarrow 16e^2-16-12=\frac{16e^2}{5}\left(e^2-1\right)$

$\Rightarrow 4e^4-24e^2+35=0$