Quadric Surfaces Six basic types of quadric surfaces: • ellipsoid • cone • elliptic paraboloid • hyperboloid of one sheet • hyperboloid of two sheets • hyperbolic paraboloid (A) (B) (C) (D) (E) (F) 1. For each surface, describe the traces of the surface in x = k, y = k, and z = k. Then pick the term from the list above which seems to most accurately describe the surface (we haven't learned any of these terms yet, but you should be able to make a good educated guess), and pick the correct picture of the surface. x2 y2 (a) − = z. 9 16 1 • Traces in x = k: parabolas • Traces in y = k: parabolas • Traces in z = k: hyperbolas (possibly a pair of lines) y2 k2 Solution. The trace in x = k of the surface is z = − 16 + 9 , which is a downward-opening parabola. x2 k2 The trace in y = k of the surface is z = 9 − 16 , which is an upward-opening parabola. x2 y2 The trace in z = k of the surface is 9 − 16 = k, which is a hyperbola if k 6= 0 and a pair of lines (a degenerate hyperbola) if k = 0. This surface is called a hyperbolic paraboloid , and it looks like picture (E) . It is also sometimes called a saddle. x2 y2 z2 (b) + + = 1. 4 25 9 • Traces in x = k: ellipses (possibly a point) or nothing • Traces in y = k: ellipses (possibly a point) or nothing • Traces in z = k: ellipses (possibly a point) or nothing y2 z2 k2 k2 Solution. The trace in x = k of the surface is 25 + 9 = 1− 4 . Since 1− 4 can be positive, zero, or negative (depending on what k is), this intersection can be an ellipse, a point (a degenerate ellipse), or nothing. x2 z2 k2 The trace in y = k of the surface is 4 + 9 = 1 − 25 , which is an ellipse, a point, or nothing. x2 y2 k2 The trace in z = k of the surface is 4 + 25 = 1 − 9 , which is an ellipse, a point, or nothing. The picture which matches this description is (A) , and this surface is an ellipsoid. x2 y2 z (c) + = . 4 9 2 • Traces in x = k: parabolas • Traces in y = k: parabolas • Traces in z = k: ellipses (possibly a point) or nothing 2y2 k2 Solution. The trace in x = k of this surface is z = 9 − 2 , which is a parabola. x2 2k2 The trace in y = k of this surface is z = 2 − 9 , which is a parabola. x2 y2 k The trace in z = k of this surface is 4 + 9 = 2 , which is an ellipse when k > 0. If k = 0, it is a point (a degenerate ellipse), and if k < 0, it is nothing. This matches picture (C) , and this surface is called an elliptic paraboloid . z2 y2 (d) − x2 − = 1. 4 4 2 • Traces in x = k: hyperbolas (never a pair of lines) • Traces in y = k: hyperbolas (never a pair of lines) • Traces in z = k: ellipses (possibly a point) or nothing z2 y2 2 Solution. The trace in x = k of this surface is 4 − 4 = 1 + k , which is a hyperbola. Since 1 + k2 is always positive, this is never a pair of lines. z2 2 k2 The trace in y = k of this surface is 4 − x = 1 + 4 , which is a hyperbola (and never a pair of lines). 2 y2 k2 k2 The trace in z = k of this surface is x + 4 = 4 − 1. Since 4 − 1 can be positive, zero, or negative, this intersection can be an ellipse, a point, or nothing. The pictures (B), (D), and (F) all show surfaces whose traces in x = k and y = k are hyperbolas and whose traces in z = k are ellipses. Of these three graphs, (D) is the only one in which some traces in z = k are nothing, so the correct picture is (D) . This surface is an example of a hyperboloid of two sheets . y2 z2 (e) x2 + = . 9 16 • Traces in x = k: hyperbolas (possibly a pair of lines) • Traces in y = k: hyperbolas (possibly a pair of lines) • Traces in z = k: ellipses (possibly a point) z2 y2 2 2 Solution. The trace in x = k of this surface is 16 − 9 = k . Since k can be either non-zero or zero, this is either a hyperbola or a pair of lines. (It is only a pair of lines when k = 0; otherwise, it is a hyperbola.) z2 2 k2 The trace in y = k of this surface is 16 − x = 9 , which is again either a hyperbola or a pair of lines (a pair of lines only when k = 0). 2 y2 k2 k2 The trace in z = k of this surface is x + 9 = 16 . Since 16 is either positive or 0, this intersection is either an ellipse or a point (but never nothing, and it is only a point when k = 0). This matches picture (F) and is an example of an cone . x2 z2 (f) + y2 − = 1. 9 16 • Traces in x = k: hyperbolas (possibly a pair of lines) • Traces in y = k: hyperbolas (possibly a pair of lines) • Traces in z = k: ellipses (never a point) 2 z2 k2 k2 Solution. The trace in x = k of this surface is y − 16 = 1 − 9 . Since 1 − 9 can be non-zero or zero, this is either a hyperbola or a pair of lines. x2 z2 2 The trace in y = k of this surface is 9 − 16 = 1 − k , which is either a hyperbola or a pair of lines. 3 x2 2 k2 k2 The trace in z = k of this surface is 9 + y = 1 + 16 . Since 1 + 16 is always positive, this is always an ellipse (and never a point or nothing). This description matches picture (B) , and this surface is a hyperboloid of one sheet . 2. Sketch the surface 9y2 + 4z2 = 36. What type of quadric surface is it? Solution. Notice that the equation describing this surface does not involve x at all. Each trace of the 2 2 y2 z2 surface in x = k is the ellipse 9y + 4z = 36, which can also be written as 4 + 9 = 1. Therefore, the surface looks like this: This is not any of the quadrics we have seen so far; it is a (quadric) cylinder . 3. Sketch the surface y2 + 2y + z2 = x2. What type of quadric surface is it? Solution. We have a y2 term as well as a y term, and we can make the equation simpler by completing the square. In this case, we accomplish that by adding 1 to both sides of the equation: y2 + 2y + 1 + z2 = x2 + 1 (y + 1)2 + z2 = x2 + 1 −x2 + (y + 1)2 + z2 = 1 To sketch, let's look at the traces: 2 2 2 • The trace in x = k of the surfacep is (y + 1) + z = 1 + k , which describes a circle (centered at y = −1, z = 0 and having radius 1 + k2). • The trace in y = k of the surface is −x2 + z2 = 1 − (k + 1)2, which is a hyperbola (or a pair of lines). • The trace in z = k of the surface −x2 + (y + 1)2 = 1 − k2, which is a hyperbola (or a pair of lines). Since we know the traces in x = k are circles centered at y = −1, z = 0, we can think of the surface as being obtained by rotating some sort of curve about the line y = −1, z = 0 (this line is parallel to the x-axis). To figure out what that curve is, we can look at the trace of the surface in either y = −1 or z = 0. Let's do the trace in z = 0: this is (y + 1)2 − x2 = 1, which is the following hyperbola: 4 y 2 1 x -3 -2 -1 1 2 3 -1 -2 -3 -4 If we draw this in space (in the plane z = 0, since we're talking about the trace in z = 0), it looks like the left figure below. Therefore, the surface looks like the right figure. z y x This is a hyperboloid of one sheet . 4. What type of quadric surface is 4x2 − y2 + z2 + 9 = 0? Solution. Let's look at the traces: • The trace in x = k of this surface is y2 − z2 = 4k2 + 9, which is always a hyperbola. • The trace in y = k of this surface is 4x2 + z2 = k2 − 9, which is an ellipse, a point, or nothing, depending on k. • The trace in z = k of this surface is y2 − 4x2 = k2 + 9, which is always a hyperbola. Since the traces in x = k and z = k are hyperbolas, we know the surface is either a hyperboloid or a cone. The fact that the traces in y = k can be nothing tells us that the surface must be a hyperboloid of two sheets. The surface looks like this: 5 6.