13.6.Ex6-14 - We rewrite the equations as follows 3 x 2 7 y...

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S E C T I O N 13.6 A Survey of Quadric Surfaces (ET Section 12.6) 399 6. x 2 3 y 2 9 z 2 = 1 SOLUTION Rewriting the equation in the form x 2 y 1 3 2 z 1 3 2 = 1 we identify it as the equation of a hyperboloid of two sheets. In Exercises 7–12, state whether the given equation defines an elliptic paraboloid, hyperbolic paraboloid, or elliptic cone. 7. z = x 4 2 + y 3 2 SOLUTION This equation defines an elliptic paraboloid. 8. z 2 = x 4 2 + y 3 2 SOLUTION This is the equation of an elliptic cone. 9. z = x 9 2 y 12 2 SOLUTION This equation defines a hyperbolic paraboloid. 10. 4 z = 9 x 2 + 5 y 2 SOLUTION The equation can be rewritten as z = x 2 3 2 + y 2 5 2 hence it defines an elliptic paraboloid. 11. 3 x 2 7 y 2 = z SOLUTION Rewriting the equation as z = x 1 3 2 y 1 7 2 we identify it as the equation of a hyperbolic paraboloid. 12. 3 x 2 + 7 y 2 = 14 z 2 SOLUTION
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Unformatted text preview: We rewrite the equations as follows: 3 x 2 + 7 y 2 = 14 z 2 3 x 2 = − 7 y 2 + 14 z 2 ⎛ ⎝ x 1 √ 3 ⎞ ⎠ 2 = − ⎛ ⎝ y 1 √ 7 ⎞ ⎠ 2 + ⎛ ⎝ z 1 √ 14 ⎞ ⎠ 2 We identify it as the equation of an elliptic cone. In Exercises 13–20, state the type oF the quadric surFace and describe the trace obtained by intersecting with the given plane. 13. x 2 + ³ y 4 ´ 2 + z 2 = 1, y = SOLUTION The equation x 2 + ( y 4 ) 2 + z 2 = 1 deFnes an ellipsoid. The xz-trace is obtained by substituting y = 0 in the equation of the ellipsoid. This gives the equation x 2 + z 2 = 1 which deFnes a circle in the xz-plane. 14. x 2 + ³ y 4 ´ 2 + z 2 = 1, y = 5...
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