The equation \(\displaystyle{x}{2}\ -\ {2}{y}{2}={4}\) can be written as:
\(\displaystyle{x}{2}\ -\ {2}{y}{2}={4}{x}{24}\ -\ {2}{y}{24}={2}{\left({x}{2}\right)}{2}\ -\ {\left({y}{2}\right)}{2}={1}\)
The standard equation of hyperbola with foci \(\displaystyle{\left(\pm\ {c},\ {0}\right)}\) will be given by:
\(\displaystyle{\left({x}{a}\right)}{2}\ -\ {\left({y}{b}\right)}{2}={1}\)
Thus, on comparing equations \(\displaystyle{x}{2}\ -\ {2}{y}{2}={4}{x}{24}\ -\ {2}{y}{24}={2}{\left({x}{2}\right)}{2}\ -\ {\left({y}{2}\right)}{2}={1}\) and \(\displaystyle{\left({x}{a}\right)}{2}\ -\ {\left({y}{b}\right)}{2}={1}\) we find that the given conic is a hyperbola.
Now, compare equation \(\displaystyle{x}{2}\ -\ {2}{y}{2}={4}{x}{24}\ -\ {2}{y}{24}={2}{\left({x}{2}\right)}{2}\ -\ {\left({y}{2}\right)}{2}={1}\) and \(\displaystyle{\left({x}{a}\right)}{2}\ -\ {\left({y}{b}\right)}{2}={1}\) and we get:
\(\displaystyle{a}={2}\ \text{and}\ {b}={2}\)
Now,
\(\displaystyle{c}={a}{2}\ +\ {b}{2}={4}\ +\ {2}={6}\)
Thus, the vertices of the given conic are \(\displaystyle{\left(\pm\ {2},\ {0}\right)}\)
The foci of the given conic is \(\displaystyle{\left(\pm\ {6},\ {0}\right)}\)
Conclusion:
Hence, the given conic section is a hyperbola. The vertices of hyperbola are \(\displaystyle{\left(\pm\ {2},\ {0}\right)}\) and the foci of the hyperbola is \(\displaystyle{\left(\pm\ {6},\ {0}\right)}\).