The locus of a point which moves such that the difference of its distances from the points $\left(\mathrm{5,0}\right)$ and $\left(-\mathrm{5,0}\right)$ is $6$ units is a conic, whose length of the latus rectum (in units) is equal to

Because $6<5-\left(-5\right)$
Hence, the locus is a hyperbola by definition.
Let's assume the eccentricity, semi-transverse axis and the semi-conjugate axis of the hyperbola is $e,a,$ and $b$ respectively, then

$2ae=10$ {Distance between focii $\left(-\mathrm{5,0}\right) \& \left(\mathrm{5,0}\right)=10=2ae$ }
& $2a=6\Rightarrow e=\frac{10}{6}=\frac{5}{3}$
$\Rightarrow 1+\frac{b^2}{a^2}=e^2=\frac{25}{9}\Rightarrow \frac{b^2}{a^2}=\frac{25}{9}-1=\frac{16}{9}$
$\Rightarrow \frac{2b^2}{a}=2\times \frac{16}{9}\times a=\frac{32}{3}$$=$ length of the latus rectum