Question

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

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

Short answer

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

Step-by-step solution

Standard form of a circular cone

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

Rewrite the given equation and compare with equation of Circular Cone

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

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

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

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

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

Graph the Circular Cone

The graph of the surface \(4{y^2} + {z^2} - x - 16y - 4z + 20 = 0\) is shown below