If the eccentricity \(e\) of the hyperbola \(\frac{{x}^{2}}{{a}^{2}}-\frac{{y}^{2}}{{b}^{2}}=1\), passing through \((6,4…

\(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\)

It passes through \((6, 4\sqrt{3})\).

\(\Rightarrow \frac{36}{a^2}-\frac{48}{b^2}=1\)\(\ldots(1)\)

Also, \(15 \mathrm{e}^2-34 \mathrm{e}+15=0\)

\(\Rightarrow 15 \mathrm{e}^2-25 \mathrm{e}-9 \mathrm{e}+15=0\)

\(\mathrm{e}=\frac{5}{3}\) or \(\frac{3}{5} \Rightarrow \mathrm{e}=\frac{5}{3}\)

\(\Rightarrow \mathrm{e}^{2}=\frac{25}{9}\)

\(1+\frac{b^2}{a^2}=\frac{25}{9} \Rightarrow \frac{b^2}{a^2}=\frac{16}{9}\) \(\ldots(2)\)

Using (1) and (2):

\(\Rightarrow \frac{36}{a^2}-\frac{48}{16 a^2} \times 9=1 \Rightarrow a=3, b=4\)

Now, the length of the latus rectum of the hyperbola \(\frac{x^2}{b^2}-\frac{y^2}{2\left(a^2+1\right)}=1\) is:

\(=\frac{4\left(a^2+1\right)}{b}=\frac{4(10)}{4}=10\)