Let us Consider the given equations of hyperbola are provided as \(\displaystyle\frac{{{x}^{{2}}}}{{9}}-\frac{{{y}^{{2}}}}{{1}}={1}{\quad\text{and}\quad}\frac{{{\left({x}-{3}\right)}^{{2}}}}{{9}}-\frac{{{\left({y}+{3}\right)}^{{2}}}}{{1}}={1}\)

The similarity between the graphs is that the distance between the vertices of both the graphs is 6 units and the difference between the graphs is that the center of the graph \(\displaystyle\frac{{{x}^{{2}}}}{{9}}-\frac{{{y}^{{2}}}}{{1}}={1}\) is the origin, since the centers of the graph \(\displaystyle\frac{{{\left({x}-{3}\right)}^{{2}}}}{{9}}=\frac{{{\left({y}+{3}\right)}^{{2}}}}{{1}}={1}{i}{s}{\left({3},-{3}\right)}\).

The similarity between the graphs is that the distance between the vertices of both the graphs is 6 units and the difference between the graphs is that the center of the graph \(\displaystyle\frac{{{x}^{{2}}}}{{9}}-\frac{{{y}^{{2}}}}{{1}}={1}\) is the origin, since the centers of the graph \(\displaystyle\frac{{{\left({x}-{3}\right)}^{{2}}}}{{9}}=\frac{{{\left({y}+{3}\right)}^{{2}}}}{{1}}={1}{i}{s}{\left({3},-{3}\right)}\).