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Chapter 13: Problem 47

Identify the quadric surface. $$ x^{2}-y^{2}+z=0 $$

Short Answer

Expert verified
The given quadric surface is a hyperbolic paraboloid.

Step by step solution

01

Identify the form of the equation

The given equation is \(x^{2}-y^{2}+z=0\). This equation can be rewritten as \(x^{2}-y^{2}=z\), indicating that the surface is symmetric with respect to the x and y axis. We can see that one variable is squared, another variable is squared but has a negative sign, and the third variable is not squared. This form suggests that the equation might represent a hyperbolic paraboloid.
02

Confirm the type of the quadric surface

To confirm that, we need to check the graph of the equation which typically exhibits a saddle shape for a hyperbolic paraboloid. But without visualization, we can still identify this based on the standard form of a hyperbolic paraboloid, which is \(z = Ax^{2}-By^{2}\) or \(z = -Ax^{2}+By^{2}\), where \(A>0\) and \(B>0\). In our equation, we can represent \(1\) as \(A\) and \(1\) as \(B\), thus confirming that our quadric surface is a hyperbolic paraboloid.

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