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Calculus IIChapter 6: 3-Dimensional SpaceSection 6-4

Quadric Surfaces — Question 8

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Question 8

A quadric has the traces z=x2z=x^2 in the plane y=0y=0 and z=−y2z=-y^2 in the plane x=0x=0. Every horizontal trace z=k≠0z=k\ne 0 is a hyperbola. Determine a simplest equation for the surface and explain whether the trace information uniquely fixes all coefficients.

Original worksheet page 1: question and worked solution for 6-4-008
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Question 8 – Solution

Strategy Combine the two vertical traces into a quadratic having the required restrictions.

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Simplest model The equation z=x2−y2\boxed{z=x^2-y^2} has exactly the stated coordinate-plane traces and horizontal hyperbolas. It is a hyperbolic paraboloid.

Uniqueness issue The given coordinate traces force the coefficients of x2x^2 and y2y^2 in a diagonal model, but without an assumption excluding a mixed term, z=x2−y2+cxyz=x^2-y^2+cxy has the same traces in x=0x=0 and y=0y=0. Its horizontal traces remain hyperbolic when the quadratic form is indefinite.

Conclusion The simplest axis-aligned equation is determined, but the data do not uniquely exclude rotated saddles.

Original worksheet page 2: question and worked solution for 6-4-008

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