13.6.Ex18-20

# 13.6.Ex18-20 - S E C T I O N 13.6 A Survey of Quadric...

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S E C T I O N 13.6 A Survey of Quadric Surfaces (ET Section 12.6) 401 18. 4 x 2 y 3 2 2 z 2 = 1, z = 1 SOLUTION We rewrite the equation as follows: x 1 2 2 y 3 2 z 1 2 2 = 1 y 3 2 z 1 2 2 + x 1 2 2 = 1 This equation defines a hyperboloid of two sheets. To find the trace of the hyperboloid on the plane z = 1, we substitute z = 1 in the equation. This gives 4 x 2 y 3 2 2 · 1 2 = 1 4 x 2 y 2 9 = 3 4 3 x 2 y 2 27 = 1 x 3 2 2 y 3 · 3 2 = 1 The trace is a hyperbola on the plane z = 1. 19. y = 3 x 2 , z = 27 SOLUTION This equation defines a parabolic cylinder, consisting of all vertical lines passing through the parabola y = 3 x 2 in the xy -plane. Hence, the trace of the cylinder on the plane z = 27 is the parabola y = 3 x 2 on this plane, that is, the following set: ( x , y , z ) : y = 3 x 2 , z = 27 . 20. y = 3 x 2 , y = 27 SOLUTION The equation
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