Question

Identify the quadric surface. $$ 2 x^{2}-y^{2}+2 z^{2}=-4 $$

Short answer

The quadric surface represented by the given equation is a hyperboloid of one sheet.

Step-by-step solution

Rewrite to standard form

Rewrite the given equation \(2 x^{2}-y^{2}+2 z^{2}=-4\) by moving the constant to the other side to result in a standard form: \(2 x^{2}-y^{2}+2 z^{2}=4\). Now we divide the entire equation by 4 in order to normalize the right side to 1: \(\frac{1}{2} x^{2} - \frac{1}{4} y^{2} + \frac{1}{2} z^{2}=1\).

Identify type of surface

Comparing the standard form equation \(\frac{1}{2} x^{2}- \frac{1}{4} y^{2} + \frac{1}{2} z^{2}=1\) to the general equations of different quadric surfaces, it matches with the equation of a hyperboloid of one sheet, which is in the form \(\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} + \frac{z^{2}}{c^{2}} = 1\), where \(a^{2} > 0\), \(b^{2} > 0\), and \(c^{2} > 0\). So, the given equation is of a hyperboloid of one sheet where \(a^{2} = 2\), \(b^{2} = -4\), and \(c^{2} = 2\).