Question

Reduce the equation to one of the standard forms, classify the surface, and sketch it.

\({x^2} - {y^2} + {z^2} - 2x + 2y + 4z + 2 = 0\)

Short answer

The surface equation \({x^2} - {y^2} + {z^2} - 2x + 2y + 4z + 2 = 0\) is a Hyperboloid.

Step-by-step solution

Standard form of a circular cone

The standard equation of hyperboloid with centre at\((h,k,l)\)along the\(y\)axis is\(\frac{{{{(x - k)}^2}}}{{{a^2}}} - \frac{{{{(y - h)}^2}}}{{{b^2}}} + \frac{{{{(z - l)}^2}}}{{{c^2}}} = 1\).

Rewrite the given equation and compare with equation of Hyperboloid

Consider the equation of the surface as\({x^2} - {y^2} + {z^2} - 2x + 2y + 4z + 2 = 0\).

Rearrange the above equation to form a standard form as follows.

\({x^2} - {y^2} + {z^2} - 2x + 2y + 4z + 2 = 0\)

\(\begin{array}{l}\left( {{x^2} - 2x + 1} \right) - \left( {{y^2} - 2y} \right) + \left( {{z^2} + 4z} \right) + 1 = 0\\\left( {{x^2} - 2x + 1} \right) - \left( {{y^2} - 2y + 1} \right) + \left( {{z^2} + 4z + 4} \right) - 4 + 2 = 0\\{(x - 1)^2} - {(y - 1)^2} + {(z + 2)^2} = 2\quad \\\frac{{{{(x - 1)}^2}}}{2} - \frac{{{{(y - 1)}^2}}}{2} + \frac{{{{(z + 2)}^2}}}{2} = 1\end{array}\)

The standard equation\(\frac{{{{(x - 1)}^2}}}{2} - \frac{{{{(y - 1)}^2}}}{2} + \frac{{{{(z + 2)}^2}}}{2} = 1\)satisfies the equation of hyperboloid, which is centred at\((1,1, - 2)\).

By the formula, the surface equation \({x^2} - {y^2} + {z^2} - 2x + 2y + 4z + 2 = 0\) is a Hyperboloid.

Graph the hyperboloid

The graph of the surface \({x^2} - {y^2} + {z^2} - 2x + 2y + 4z + 2 = 0\) is shown below in Figure