Equation of the hyperbola of eccentricity 3 and the distance between whose foci is 24 is

Given $dis\tan ce between foci=2ae=24 and \mathrm{e}=3$

$\Rightarrow ae=12$

$$\begin{gathered} \Rightarrow \mathrm{a}\left(3\right)=12 \\ \Rightarrow \mathrm{a}=4 \\ \mathrm{Now} {\mathrm{b}}^2={\mathrm{a}}^2({\mathrm{e}}^2-1)=16\left(8\right)=128 \\ \mathrm{Hyperbola} \mathrm{is} \frac{{\mathrm{x}}^2}{16}-\frac{{\mathrm{y}}^2}{128}=1 \\ \Rightarrow 8{\mathrm{x}}^2-{\mathrm{y}}^2=128 \end{gathered}$$