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 xztrace 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 xzplane. 14. x 2 + ³ y 4 ´ 2 + z 2 = 1, y = 5...
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 Winter '08
 GANGliu
 Hyperboloid, Ellipsoid, elliptic cone

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