Exercise · ZW-QUADRIC-SURFACE-005 · ★☆☆ Basic
What is the intersection of \(x^2+y^2-z^2=-1\) with the plane \(z=0\)?
Choose one option.
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- B. Empty
Explanation
With \(z=0\) we get \(x^2+y^2=-1\), which has no real solutions. That is why the surface has two sheets, one for \(z\ge1\) and one for \(z\le-1\).
Review the concept: Quadric surface
A quadric surface is the set of points \((x,y,z)\) in space that satisfy a second-degree equation in \(x\), \(y\), \(z\) with real coefficients \(A,B,C,D,E,F,G,H,I,J\).
\[Ax^2+By^2+Cz^2+Dxy+Exz+Fyz+Gx+Hy+Iz+J=0\]
More practice problems: quadric surface
- ZW-QUADRIC-SURFACE-001 Find the radius of the surface x^2+y^2+z^2=9.
- ZW-QUADRIC-SURFACE-002 Which quadric is (x^2)/(4)+(y^2)/(9)+z^2=1?
- ZW-QUADRIC-SURFACE-003 What curve results from cutting z=x^2+y^2 with the plane z=4?
- ZW-QUADRIC-SURFACE-004 Classify x^2+y^2-z^2=1.
- ZW-QUADRIC-SURFACE-006 What is the trace of z^2=x^2+y^2 in the plane z=0?
- ZW-QUADRIC-SURFACE-007 For z=x^2-y^2, what is the trace in the plane x=0?
- ZW-QUADRIC-SURFACE-008 Complete the square in x^2+y^2+z^2-2x+4y=4 and give the center and radius.
- ZW-QUADRIC-SURFACE-009 Which of these points lies on the surface x^2+y^2=4 in space?