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3 3.1 Plane quadratic curves 59 It turns out to be convenient to rewrite this expression using matrices and the column form of vectors. Here is the rewritten expression: ( ) ( ) ( ) x1 cos α x 2 sin α cos α sin α x1 =, x 1 sin α + x 2 cos α sin α cos α with a matrix product on the right hand side (see below). Here are the details: A matrix is a rectangular array of (possibly variable) numbers surrounded by a pair of brackets. It is a way to store (mathematical) information, which has proven its usefulness. In a way it is comparable to a table for collecting data. A matrix contains a number of rows and a number of columns. An r by s matrix is a matrix with r rows and s columns. Here is an example of a 2 by 3 matrix: ( ) 2 x 2 5 sin α 0 2 t The i, j th element of the matrix is the element which is in the i th row and in the j th column. In the example, the 1, 3 th element is 5 sin α. The left hand side contains the vector (x 1 cos α x 2 sin α, x 1 sin α + x 2 cos α) written in column form. This is a special case of a matrix: a 2 by 1 matrix. The right hand side contains the product of two matrices: a 2 by 2 matrix containing cosines and sines and a 2 by 1 matrix containing the column form of the vector (x 1, x 2 ). The result of the multiplication is the 2 by 1 matrix (vector in column form) on the left hand side. The matrix multiplication works as follows. The product is a 2 by 1 matrix whose first entry is obtained by taking the (2d) inner product of the first row and the vector (x 1, x 2 ), i.e., (cos α, sin α) (x 1, x 2 ) = x 1 cos α x 2 sin α. The second entry is obtained similarly, by taking the inner product of the second row of the 2 by 2 matrix and the vector (x 1, x 2 ): (sin α, cos α) (x 1, x 2 ) = x 1 sin α + x 2 cos α. So the rotation over α (radians in our case) is encoded by the 2 by 2 matrix ( ) cos α sin α x 2 sin α cos α in the sense that multiplying this matrix with the column form of the vector (x 1, x 2 ) produces the column form of the rotated vector. The way to remember this rotation matrix is the following:

4 60 Quadratic curves, quadric surfaces The first column contains the result of rotating (1, 0) over α radians. The second column contains the result of rotating (0, 1) over α radians. Here are a few examples. The matrix for a rotation over π/6 or 30 is To rotate the vector (4, 2), we multiply as follows 3 1 ( ) ( = ) = ( ). So (4, 2) transforms into (2 3 1, 2 + 3). The matrix describing a rotation over α is ( ) cos α sin α. sin α cos α We will have more to say on matrices and their use later A brief word on handling quadratic equations We briefly address the question of how to handle a quadratic equation in order to get some idea of what shape the equation represents. There are two types of operations which are vital. Splitting off squares Whenever an equation contains a square and a linear term in the same variable, like 2x 2 12x, these two terms can be rewritten as follows: 2x 2 12x = 2(x 2 6x) = 2((x 3) 2 9) = 2(x 3) This technique is known as spliting off a square.. Using this technique, the equation x 2 + 6x + y 2 + 4y = 23 can be rewritten as (x + 3) (y + 2) 2 4 = 23, and finally as (x + 3) 2 + (y + 2) 2 = 36. This represents a circle with center ( 3, 2) and radius 6 (see below for more on the circle). As you see, the technique of splitting off a square is related to translations.

5 3.1 Plane quadratic curves 61 y y x x Figure 3.2: The left hand ellipse turns out to have a relatively simple equation without a mixed term xy. The right hand side is an ellipse whose equation does contain a mixed term xy. By applying a suitable rotation around the origin, in this case over 45, the equation of the rotated figure simplifies and can then be recognized as an ellipse. Getting rid of a mixed term xy Whenever a quadratic equation contains (a multiple of) the mixed product xy, a suitable rotation can be used to get rid of this term. For example, consider the equation x 2 + xy + y 2 = 9. Suppose we rotate the corresponding figure over an angle of α radians. If (u, v) is on the rotated figure, then rotating it over α radians produces a point on x 2 + xy + y 2 = 9. By the intermezzo, the rotation changes (u, v) into (u cos α v sin α, u sin α + v cos α). The problem is to find α so that the equation satisfied by this point is simple. So we proceed to substitute the coordinates in the equation: (u cos α v sin α) 2 + (u cos α v sin α)(u sin α + v cos α) + (u sin α + v cos α) 2 = 9. Expanding the expressions on the left hand side and using the identity cos 2 α + sin 2 α = 1 yields the equation: (1 + sin α cos α)u 2 + (1 sin α cos α)v 2 + uv(cos 2 α sin 2 α) = 9. The term uv(cos 2 α sin 2 α) on the left hand side is the crucial one: if we choose α in such a way that sin 2 α = cos 2 α, the whole term vanishes! This is, for example, 2 the case if we choose α = π/4 radians (or 45 ), since then cos α = sin α = 2. The equation reduces to 3 2 u v2 = 9 or 3u 2 + v 2 = 18. Below you will learn that this equation describes an ellipse Listing all types of quadratic curves Using techniques like the above, one can unravel the structure of any quadratic curve. These are roughly the steps.

6 62 Quadratic curves, quadric surfaces a) First remove, by a suitable rotation, a mixed term cxy. b) In the next step, the linear terms can be dealt with. Two cases need to be distinguished. If in the equation an expression occurs like ax 2 + bx (with a 0), then a square can be split off. If a variable, say x, occurs in a linear term, but x 2 does not occur in the equation, then we can not split off a square, and the linear term can not be removed by a suitable translation. For instance, the equation x 2 +2x 2y = 3 can be rewritten as 2y = (x + 1) 2 4. The term 2y survives so to speak. The equation represents a parabola. In the end, one is left with an equation containing x 2, y 2 or both, containing at most one linear term, and a constant. For instance, 2x 2 y 2 = 1 or x 2 y = 0. The list is then found by enumerating the possibilities: two squares with positive coefficients (like 2x 2 +3y 2 = 1), two squares precisely one of whose coefficients is positive (like x 2 2y 2 = 2), one square and a linear term (like x 2 2y = 5). Here are the details. The circle The circle The circle with center (0, 0) and radius r has the well known equation x 2 + y 2 = r 2. If you translate the circle over the vector t = (a, b), then the new cirle will have equation (x a) 2 + (y b) 2 = r 2. This can be seen by using the method that we introduced in Section 2.6 of Chapter 2: if (u, v) is on the translated circle, then (u a, v b) is on the circle x 2 + y 2 = r 2. So (u a, v b) satisfies (u a) 2 + (v b) 2 = r 2. The circle is an extremely symmetric curve: rotating it over any angle around its center r (a,b) Figure 3.3: Circle with center (a, b) and radius r. produces a copy that coincides with the original circle. Only individual points have actually moved to a new position. In geometric terms, a circle consists of all points with a fixed distance to a given point. In vector terms, the circle with center p and radius r is described by x p = r.

7 3.1 Plane quadratic curves 63 The ellipse The standard equation of an ellipse The circle is a special kind of ellipse. The standard form of the equation of an ellipse is x 2 a 2 + y2 b 2 = 1, where a and b are non zero constants. By convention, these are taken to be positive. If a = b, then the ellipse is a circle with radius a. You could view the ellipse as a kind of circle which is being stretched in the x and y direction. The numbers a and b determine the amount of stretching. You may wonder why people prefer this standard form over, for instance, a 2 x 2 +b 2 y 2 = 1. One reason is that with the first equation, the points of intersection with the x axis and the y axis look relatively simple: the points of intersection with the x axis are (a, 0) and ( a, 0); the points of intersection with the y axis are (0, b) and (0, b). (If a > b, then the segment connecting ( a, 0) and (a, 0) is called the long axis of the ellipse, and the segment connecting ( b, 0) and (b, 0) the short axis). Anyway, the equation a 2 x 2 +b 2 y 2 = 1 describes an ellipse just as well The geometric description of an ellipse An ellipse can also be described by a geometric property, quite similar to the geometric description of a circle. However, the description is more involved. Here is this geometric P F F 1 2 Figure 3.4: Ellipse with foci F 1 en F 2 (left). The points on the ellipse, like P, satisfy P F 1 + P F 2 = r, for some constant r > F 1 F 2. The figure on the right hand side illustrates the long and short axes of the ellipse. Both axes are (if a b the only) axes of symmetry of the ellipse. description of an ellipse. Take two points in the plane, say F 1 and F 2. Fix a positive number r which is greater than the distance between F 1 and F 2. The set of points P satisfying P F 1 + P F 2 = r is called an ellipse with foci F 1 and F 2.

8 64 Quadratic curves, quadric surfaces Here is how you get from this description to the (standard form) of the equation of an ellipse. Take the x axis through the foci, and take the perpendicular bisector of F 1 F 2 as y axis. Suppose F 1 = ( c, 0) and F 2 = (c, 0). We write the defining property as P F 1 + P F 2 = 2a, where a is a positive constant with a > c (the factor 2 is chosen so that we arrive at the standard form in the end; just wait and see). Using Pythagoras Theorem in the triangles P =(x,y) F O 1 F 2 Q ( c,0) (c,0) Figure 3.5: Analysing the condition P F 1 + P F 2 = 2a. Use Pythagoras theorem in the triangles P F 1 Q and P F 2 Q. For instance, P F 1 = F 1 Q 2 + P Q 2 = (x + c) 2 + y 2. P F 1 Q and P F 2 Q the equality P F 1 + P F 2 = 2a becomes (x + c)2 + y 2 + (x c) 2 + y 2 = 2a. Here are the steps that lead you to the standard equation. Bring one of the square roots to the right hand side and square both sides of the resulting equation: (x c) 2 + y 2 = (2a (x + c) 2 + y 2 ) 2. Expand the right hand side and bring all terms except the term 4a (x + c) 2 + y 2 to the left hand side. Square both sides of the resulting equation. A bit of rewriting, which we leave out here, produces the equation (a 2 c 2 ) x 2 + a 2 y 2 = a 2 (a 2 c 2 ). If we take b = a 2 c 2, then the equation becomes b 2 x 2 + a 2 y 2 = a 2 b 2. Dividing both sides by a 2 b 2 finally yields x 2 a 2 + y2 b 2 = 1.

9 3.1 Plane quadratic curves Translating and rotating ellipses The effect of a translation over t = (r, s) of the ellipse with equation x2 a + y2 = 1 is an 2 b 2 ellipse whose equation is (x r) 2 (y s)2 + = 1, a 2 b 2 as can be verified just like in the case of a circle. The effect of a rotation is more involved: for most rotations, the result is an equation which is not easily recognized anymore as the equation of an ellipse. Here is one example. Rotate the ellipse over 90 in the positive direction, i.e., every point (x, y) is transformed into ( y, x). If (u, v) is on the rotated ellipse, then rotate it back over 90. The resulting point is (v, u) and lies on the ellipse we started with, so satisfies the equation So the rotated ellipse has equation v 2 a 2 + ( u)2 b 2 = 1. x 2 b 2 + y2 a 2 = 1, i.e., the roles of a and b have been interchanged. The parabola Parabolas The standard form of the equation for a parabola is y = ax 2, where a is a non zero constant. There is also a geometric description, which starts from a point F (the focus in the plane, and a line l not through F. A parabola with focus F and directrix l consists of the points P in the plane satisfying P F = distance of P to l. Like in the case of an ellipse, the equation of a parabola can be derived from this geometric description. If a mirror is designed in the shape of a parabola, then light rays are reflected through the focus. The hyperbola Hyperbolas The standard equation of a hyperbola is x 2 a 2 y2 b 2 = 1,

10 66 Quadratic curves, quadric surfaces 5 4 P 3 2 P F F directrix Figure 3.6: The defining property of a parabola (left): the distance of a point P on the parabola to F is equal to its distance to the directrix. For focus F = (0, 1/2) and direction line y = 1/2, the points P on the parabola have equal distance to F and to the line. This translates into x 2 + (y 1/2) 2 = y + 1/2. Simplifying yields the equation y = x 2 /2 (right). P 3 2 P F1 F1 F F2 2 3 Figure 3.7: The defining property of a hyperbola (left): the absolute value of the difference of the distances of a point P on the curve to the foci F 1 and F 2 is constant. In the case F 1 = ( 2, 0), F 2 = ( 2, 0), the equation P F 1 P F 2 = 2 translates into x 2 y 2 = 1 (right). where a and b are, by convention, positive constants. The hyperbola is special in the sense that it consists of two separate branches. Again there is a geometric definition of a hyperbola, this time involving two foci F 1 and F 2. Given these points and a positive constant d, the set of points P such that the absolute value of the difference of the distances P F 1 and P F 2 is equal to d, i.e., P F 1 P F 2 = d is by definition a hyperbola. Just as in the case of an ellipse or a parabola, an equation of a hyperbola can be derived from this definition. In fact, if the foci are ( a 2 + b 2, 0) and ( a 2 + b 2, 0), and the constant is 2a (with a and b positive), then it can be shown that the equation of the corresponding hyperbola is the standard equation x2 a 2 y2 b 2 = 1.

11 3.2 Parametrizing quadratic curves 67 Remaining types: degenerate quadratic curves Degenerate quadratic curves Strictly speaking there are two more types of quadratic curves: a) A pair of distinct lines. This occurs when the quadratic expression q(x, y) in the equation q(x, y) = 0 can be written as a product of two linear factors: q(x, y) = l(x, y)m(x, y). For example, x 2 y 2 = 0 is such an equation. It can be rewritten as (x + y)(x y) = 0. So (x, y) satisfies this equation precisely if it satisfies x + y = 0 (a line) or x y = 0 (another line). Again, it may not always be clear from a given quadratic equation, whether it can be written as a product of two linear factors. b) The second case is even more special: suppose q(x, y) in the equation q(x, y) = 0 is actually a square of a linear factor, say q(x, y) = l 2 (x, y). Then q(x, y) = 0 is equivalent with l(x, y) = 0. So the equation represents a line. (Mathematicians say a line with multiplicity 2.) 3.2 Parametrizing quadratic curves Parametric descriptions versus equations In Chapter 2 we described planes in 3 space in two ways: with equations and with parametric descriptions. The first way defines the points on a plane in an implicit way (you have to solve the equation to get the actual coordinates of points), the second way provides you with explicit coordinates of the points, once you substitute actual values for the parameters. Both approaches have their advantages. In this section we discuss parametric descriptions of quadratic curves. Although the various types of curves admit more parametric descriptions, the best known parametric descriptions are derived from the two identities cos 2 u + sin 2 u = 1 and cosh 2 u sinh 2 u = Parametrizing circles and ellipses A useful parametric description of the circle x x 2 2 = r 2 (with r > 0) is based on the identity cos 2 φ + sin 2 φ = 1. Here it is: x 1 = r cos φ x 2 = r sin φ. The parameter φ has a clear geometric meaning: it is the angle between the vectors (1, 0) and x = (x 1, x 2 ), and runs, say, from 0 to 2π. For instance, the circle x x 2 2 = 16 has parametric description (4 cos φ, 4 sin φ). For φ = 30 we get the point 3 (4 2, ) = (2 3, 2).

12 68 Quadratic curves, quadric surfaces To deal with the ellipse with equation x2 1 a + x2 2 2 b 2 parametric description of the circle works: x 1 = a cos φ x 2 = b sin φ, = 1, we note that a small change of the with φ ranging from, say, 0 to 2π. Here is how you check that this parametric description really satisfies the equation: (a cos φ) 2 (b sin φ)2 + = a2 cos 2 φ + b2 sin 2 φ = cos 2 φ + sin 2 φ = 1. a 2 b 2 a 2 b 2 In the parametric description x 1 = a cos φ, x 2 = b sin φ, the parameter φ does not have such a clear geometric meaning as in the case of a circle. Some other ways to parametrize circles and ellipses are discussed in the exercises Parametrizing hyperbolas Let us first look at the equation x 2 1 x 2 2 = 1, representing a relatively simple hyperbola. This equation contains a difference of two squares. In parametrizing a sum of two squares we used the cosine and the sine, because the sum of their squares is 1. Now maybe you recall from previous mathematics courses that there also exist two functions such that the difference of their squares equals 1, the hyperbolic cosine and the hyperbolic sine: cosh 2 u sinh 2 u = 1. This leads to the following parametric description for the hyperbola x 2 1 x 2 2 = 1: x 1 = cosh u x 2 = sinh u, where u runs through the real numbers. This parametric description is easily adapted to one for the standard equation x2 1 a 2 x2 2 b 2 = 1: x 1 = a cosh u x 2 = b sinh u Unfortunately, these parametric descriptions only catch one half of the hyperbola: since cosh u > 0 for all u, no points of the left hand branch of the hyperbola are parametrized. For the hyperbola, we need a separate parametric description for the left hand branch, for instance: x 1 = cosh u x 2 = sinh u, where u run through the real numbers Parametrizing parabolas The standard parabola x 2 = x 2 1 is of the form y = f(x), where f : R R is the function with f(x) = x 2. So, it is easy to parametrize the parabola as follows: x 1 = u x 2 = u 2.

15 3.3 Quadric surfaces Intermezzo: Rotations around coordinate axes in R 3 A rotation of x = (x 1, x 2, x 3 ) around the z axis over α radians does not affect the third coordinate, x 3. But the change in x 1 and x 2 is like the change in rotating (x 1, x 2 ) in the x 1, x 2 plane around the origin: so (x 1, x 2, x 3 ) is rotated to (x 1 cos α x 2 sin α, x 1 sin α + x 2 cos α). As in the planar case, it turns out to be more convenient to use matrices. So we write the vectors in column form (a 3 by 1 matrix) and use a matrix product to describe the rotation: cos α sin α 0 sin α cos α x 1 x 2 x 3 = x 1 cos α x 2 sin α x 1 sin α + x 2 cos α x 3 The left hand side contains a 3 by 3 matrix containing all information regarding the rotation. Similar to the 2d case, the first column contains the result of rotating e 1 = (1, 0, 0), the second column the result of rotating e 2 = (0, 1, 0) and the third column the result of rotating e 3 = (0, 0, 1). Here is how the matrix product works: The product of a 3 by 3 matrix and a 3 by 1 matrix (like a vector in column form) is a 3 by 1 matrix (which appears on the right hand side of the = sign). The first entry of the matrix product is obtained by taking the inner product of the first row of the 3 by 3 matrix and the vector. In our case: (cos α, sin α, 0) (x 1, x 2, x 3 ) = x 1 cos α x 2 sin α. Likewise, the second entry is obtained by taking the inner product of the second row of the 3 by 3 matrix and the vector: (sin α, cos α, 0) (x 1, x 2, x 3 ) = x 1 sin α + x 2 cos α. Finally, the inner product of the third row and the vector yields the third entry: (0, 0, 1) (x 1, x 2, x 3 ) = x 3. Rotations around the other coordinate axes can be dealt with in a similar manner. For instance, a rotation around the x 1 axis over α radians leaves the first coordinate of x = (x 1, x 2, x 3 ) fixed but changes the second and third coordinates: x (x 1, x 2 cos α x 3 sin α, x 2 sin α + x 3 cos α). In matrix form, this becomes: cos α sin α 0 sin α cos α x 1 x 2 x 3 = x 1 x 2 cos α x 3 sin α x 2 sin α + x 3 cos α For instance, the second entry on the right hand side is obtained as the inner product (0, cos α, sin α) (x 1, x 2, x 3 ) = x 2 cos α x 3 sin α.

16 72 Quadratic curves, quadric surfaces Spheres Probably the most familiar quadric surface is the sphere, the collection of points or vectors at a fixed distance from a fixed point or vector (the center). In cartesian coordinates, a sphere with center at (0, 0, 0) and radius r has equation x 2 + y 2 + z 2 = r 2. The sphere is extremely symmetric: no matter how you rotate the sphere around its center, the result will always coincide with the original sphere. But if you move the sphere so that its center is at (a, b, c), then the equation does change of course. If (u, v, w) is on the new sphere, then (u a, v b, w c) is on x 2 + y 2 + z 2 = r 2, so that (u, v, w) satisfies (u a) 2 + (v b) 2 + (w c) 2 = r 2. In conclusion, the equation of the sphere with center at (a, b, c) and radius r is (x a) 2 + Figure 3.10: A sphere and an ellipsoid. (y b) 2 + (z c) 2 = r 2 (note the minus signs!) Ellipsoids A sphere is a special kind of ellipsoid, more or less like a circle is a special kind of ellipse. The equation x 2 a + y2 2 b + z2 2 c = 1 2 (by convention a, b and c are all taken to be positive: a > 0, b > 0, c > 0) represents an ellipsoid with semi axes (of length) a, b, and c. In the special case a = b = c, the ellipsoid simplifies to a sphere with radius a. The presence of the parameters a, b, and c, linked to the three coordinate variables x, y, and z, sort of breaks the symmetry present in the sphere. Some of the symmetry is restored if, for instance, a = b. If you intersect the ellipsoid x 2 a 2 + y2 a 2 + z2 c 2 = 1

17 3.3 Quadric surfaces 73 with the plane z = c/2, a glimpse of this symmetry becomes apparent: Within the plane z = c/2, the equation of the resulting curve is x 2 a 2 + y2 a 2 = 3 4, representing a circle with radius a 3 2. Translating the ellipsoid over t = (t 1, t 2, t 3 ) produces an ellipsoid with equation (x t 1 ) 2 a 2 + (y t 2) 2 b 2 + (z t 3) 2 c 2 = 1. Since our ellipsoid lacks rotational symmetry in general, rotating it will produce a surface with a different equation. For instance, rotating around the z axis over 90 according to (x, y, z) ( y, x, z) (how does this follow from our discussion on rotations?) will produce the following change. if (u, v, w) is on the rotated ellipsoid, then rotating it back yields (v, u, w), which must be on our ellipsoid. So (v, u, w) must satisfy i.e., the equation of the rotated ellipsoid is v 2 a + ( u)2 + w2 2 b 2 c = 1, 2 x 2 b 2 + y2 a 2 + z2 c 2 = 1. Note that the effect of the rotation on the equation is that the coefficients a and b have been interchanged. Other rotations may change the equation so drastically, that from the new equation it is much harder to recognize the ellipsoid Cylinders The equation x 2 + y 2 = r 2 represents a (right circular) cylinder of radius r, whose so called axis is the z axis. Note that this equation is independent of z. This means that, given any point on the cylinder, the whole vertical line through that point is also on the cylinder. Translating the cylinder over t = (t 1, t 2, t 3 ) changes the equation in the following, by now straightforward, way: (x t 1 ) 2 + (y t 2 ) 2 = r 2. (Note that t 3 does not occur. Can you explain this geometrically?) Of course, the cylinder x 2 + y 2 = r 2 has circular symmetry with respect to the z axis: if you turn the cylinder around the z axis over any angle, the result will coincide with the original cylinder (although individual points have moved of course to another place on the

18 74 Quadratic curves, quadric surfaces cylinder). Here is the algebraic verification. By the intermezzo, a rotation over an angle α is described by the transformation (x, y, z) (x cos α y sin α, x sin α + y cos α, z). Rotating backwards, i.e., rotating over an angle α, is then described by (x, y, z) (x cos α+y sin α, x sin α+y cos α, z) (where we have used cos( α) = cos α and sin( α) = sin(α)). If (u, v, w) is on the rotated cylinder, then turning it back produces (u cos α + v sin α, u sin α + v cos α, z). These coordinates should satisfy the equation x 2 + y 2 = r 2 : (u cos α + v sin α) 2 + ( u sin α + v cos α) 2 = r 2. Now this looks like an awful expression, but see what happens to the left hand side, mainly thanks to the fact that cos 2 α + sin 2 α = 1: (u cos α + v sin α) 2 + ( u sin α + v cos α) 2 = u 2 cos 2 α + 2u cos α v sin α + v 2 sin 2 α + ( u) 2 sin 2 α 2u sin α v cos α + v 2 cos 2 α = u 2 (cos 2 α + sin 2 α) + v 2 (cos 2 α + sin 2 α) = u 2 + v 2. Conclusion: the rotated cylinder has the same equation, x 2 + y 2 = r 2. Probably you agree that it is quite remarkable, that such an obvious geometric fact involves such a lengthy computation. The equation x 2 a + y2 2 b = 1 2 (with, by convention, a > 0 and b > 0) represents a cylinder whose horizontal cross section is an ellipse. If a = b, then this equation reduces to x 2 + y 2 = a 2, i.e., a circular cylinder of radius a. You can also view the cylinder as being constructed in the following way. Start Figure 3.11: On the left, the cylinder x 2 +y 2 = 1. On the right the cylinder (x z) 2 +y 2 = 1. with a circle, say x 2 +y 2 = 1 in the plane z = 0. Now the vertical lines passing through the circle together form the cylinder. Given this construction you can wonder what happens

19 3.3 Quadric surfaces 75 if instead of vertical lines, you take lines with a fixed direction which is not vertical, as illustrated in the figure. We will discuss such an example when we deal with parametric descriptions of quadrics. Instead of considering cylinders over circles, you could just as well consider cylinders over other quadratic curves, like a parabola. For example, the equation y = x 2 describes a right parabolic cylinder Cones The equation x 2 + y 2 = z 2 describes a right circular cone with the z axis as vertical axis. A horizontal cross section at level z = r produces the circle x 2 + y 2 = r 2 (and just a point for r = 0). A vertical Figure 3.12: Part of the cone x 2 +y 2 z 2 = 0 (left) and the cone x 2 +y 2 z 2 /4 = 0 (right). The coefficient of z 2 controls the steepness. cross section, say with the plane y = 0, produces a pair of lines in the plane y = 0: x 2 = z 2 (and y = 0), which reduces to x z = 0 and y = 0, or x + z = 0 and y = 0. In particular, a cone contains lines. The vertical cross section with the plane y = r for r 0 is a hyperbola: x 2 z 2 = r 2. By introducing parameters in the equation, the shape can be adapted. For example, the parameter a in x 2 + y 2 = a 2 z 2 increases or decreases the steepness of the cone. The standard equation of the general cone is given by x 2 a 2 + y2 b 2 = z2 c 2. It has less symmetry than our right circular cone. The cone also arises as follows: take a circle at level z = 1, say x 2 + y 2 = 1. Now connect each point of this circle by a line with the origin. These lines then run through a cone. To see this, take an arbitrary point on the circle, say (cos u, sin u, 1) for some

20 76 Quadratic curves, quadric surfaces real number u. The line through the origin and this point is given by the parametric description λ (cos u, sin u, 1). So any vector on the surface is described by some vector of the form x(u, λ) = λ (cos u, sin u, 1). These vectors satisfy the equation of the cone: (λ cos u) 2 + (λ sin u) 2 λ 2 = λ 2 (cos 2 u + sin 2 u) λ 2 = λ 2 λ 2 = Paraboloids This type is related to the parabola. We distinguish two types: The elliptic paraboloid. The equation (in standard form) z = x2 a + y2 2 b 2 describes such a surface. The intersection of this quadric with a vertical plane, like the plane y = 0 (or y = c, any constant) yields a parabola, hence the name paraboloid. By intersecting with a horizontal plane, like z = 1, we get an ellipse. This explains the adjective elliptic. Figure 3.13: An elliptic paraboloid (left) and a hyperbolic paraboloid (right). The hyperbolic paraboloid. Its equation in standard form is z = x2 a y2 2 b 2 (note the minus sign). The intersection with a horizontal plane consists of a hyperbola. For instance, for z = 1, we get the hyperbola x2 a y2 = 1 (in the plane z = 1!). 2 b2 The intersection with a vertical plane produces a parabola Hyperboloids The hyperboloids come in two types, a one sheeted and a two sheeted version.

21 3.3 Quadric surfaces 77 Figure 3.14: Part of the hyperboloid x 2 + y 2 z 2 = 1 (left). The right hand side illustrates what happens when the equation changes from x 2 + y 2 z 2 = 1 (hyperboloid of one sheet) to x 2 + y 2 z 2 = 0 (cone), and finally to x 2 + y 2 z 2 = 1 (hyperboloid of two sheets). a) The standard equation of a hyperboloid of one sheet is x 2 a + y2 2 b z2 2 c = 1, 2 where a, b and c are positive constants. We illustrate the fact that this type of hyperboloid consists of one piece (one sheet) by taking a closer look at the hyperboloid with a = b = c = 1, i.e., x 2 +y 2 z 2 = 1. If you rewrite the equation as x 2 +y 2 = 1+z 2, then you clearly see that the intersection with any horizontal plane z = constant produces a circle. For instance, for z = 0, we get the circle x 2 + y 2 = 1 in the plane z = 0. By varying the position of these horizontal planes, you find a varying family of circles. The intersection with a vertical plane usually consists of a hyperbola. For instance, the plane x = 0 meets x 2 + y 2 z 2 = 1 in the hyperbola with equation y 2 z 2 = 1 (in the plane x = 0). b) The standard equation of the hyperboloid of two sheets is x 2 a + y2 2 b z2 2 c = 1, 2 with positive constants a, b and c. Note that the only difference with the previous equation is the minus sign on the right hand side. To explain the two sheetedness, we take a closer look at x 2 + y 2 z 2 = 1. For z < 1, this equation has no solutions, i.e., between the horizontal planes z = 1 and z = 1, there is no point of the hyperboloid. But the intersection of the figure with the plane z = 3, for example is again a circle. In general, the cross section with a horizontal plane z = c with c 1 produces a circle. All these cross sections together make up one sheet. The cross sections with the planes z = c with c 1 make up the other sheet Degenerate quadrics As with quadratic curves, some quadric surfaces are more or less degenerate. Here are typical examples.

23 3.4 Parametrizing quadrics 79 squares instead of two squares. To relate this situation to a sum of two squares, we rewrite the equation as ( ) 2 x x x 2 3 = 1. Now, this is a sum of two squares, where the square root also contains a sum of two squares. We then proceed in two steps: a) For a given point (x 1, x 2, x 3 ) satisfying the equation, choose u between 0 and π so that x x 2 2 = sin u and x 3 = cos u. (Why is this possible? Think of the parametric description of a circle.) z u y x v Figure 3.15: Spherical coordinates (r cos v sin u, r sin v sin u, r cos u) with 0 u π and 0 v 2π are a widely used way of parametrizing a sphere with center (0, 0, 0) and radius r. The parameters have a clear geometric meaning. If x = (r cos v sin u, r sin v sin u, r cos u), then u is the angle between x and the positive z axis, and v is the angle between the positive x axis and the projection (x 1, x 2, 0) of x on the x, y plane. The horizontal and vertical curves in the picture on the right hand side are curves where either u or v is kept constant. b) Now parametrize, for fixed u, the equation x x 2 2 = sin 2 u in the standard way: x 1 = cos v sin u, x 2 = sin v sin u. (With v between 0 and 2π.) Collecting the two steps leads to the following parametrization of the sphere x x x 2 3 = 1: x 1 = cos v sin u, x 2 = sin v sin u, x 3 = cos u. This parametric description also goes by the name parametric description in spherical coordinates. Both u and v have a clear geometric meaning: u is the angle between e 3 = (0, 0, 1) and x, and v is the angle between e 1 and the vector (x 1, x 2, 0). From here on, it is fairly easy to parametrize the ellipsoid x2 1 a 2 + x2 2 b 2 + x2 3 c 2 = 1: x 1 = a cos v sin u, x 2 = b sin v sin u, x 3 = c cos u.

24 80 Quadratic curves, quadric surfaces Parametrizing hyperboloids To parametrize the hyperboloid of one sheet with equation x x 2 2 x 2 3 = 1, we will need both (co)sines and hyperbolic (co)sines to deal with the plus and minus signs. First, we rewrite our equation as a difference of squares: ( ) 2 x x 2 2 x 2 3 = 1, as in the case of a sphere. Then we find a parametrization in the following two steps: a) First we handle the difference of squares with hyperbolic (co)sines: x x 2 2 = cosh u, x 3 = sinh u. b) For fixed u, we then parametrize the equation x x 2 2 = cosh 2 u in the usual way: x 1 = cosh u cos v, x 2 = cosh u sin v. The resulting parametric description is x 1 = cosh u cos v, x 2 = cosh u sin v, x 3 = sinh u. To parametrize (one sheet of) a hyperboloid of two sheets, say with equation x 2 1 +x 2 2 x 2 3 = 1, the first option is to rewrite the equation in the following way as a difference of two squares: ( 2 x 2 3 x x2) 2 = 1. Next, we handle the difference of two squares with x 3 = cosh u, x x 2 2 = sinh u. For fixed u we parametrize the circle x x 2 2 = sinh 2 u with x 1 = sinh u cos v and x 2 = sinh u sin v. The resulting parametrization x 1 = sinh u cos v, x 2 = sinh u sin v, x 3 = cosh u describes only the part of the hyperboloid above the x 1, x 2 plane. Do you see why? Parametrizations and curves on quadrics With a parametric description of a quadric come two families of curves on the quadric. These families of curves can be visualized as special patterns on the quadric. Consider, for example, the cylinder x 2 + y 2 = 1 parametrized by x = cos u, y = sin u, z = v (with 0 u 2π and < v < ). If we keep u fixed, say u = π, and let v run through the real numbers, the result is a vertical line on the cylinder. If, on the other hand, we keep v fixed and let u run through the interval [0, 2π], we obtain a circle at height v on the cylinder. A different parametrization usually results in a different pattern on the cylinder as is illustrated in the figure. If we keep u constant in the parametrization x = cos(u + v/2),

26 82 Quadratic curves, quadric surfaces does not change its position. In the following we only need to consider the case of a plane at distance at most 1 from the origin. If the distance to the plane is greater than 1, then the plane does not meet the sphere, since all points of the sphere are at distance 1 from the origin. a) Suppose we have a plane at distance r from the origin with 0 r 1. Rotate the plane around the origin, so that the rotated plane is horizontal (parallel with the x, y plane) and above the x, y plane (or coincides with it, for r = 0). Its equation will be z = r. b) Now the intersection of z = r and x 2 + y 2 + z 2 = 1 is a circle in the plane z = r with equation x 2 + y 2 = 1 r 2. If r = 1, then we get precisely one point (the point (0, 0, 1), the north pole ). If r = 0, we get a circle with radius 1 in the x, y plane: x 2 + y 2 = 1 and z = Intersecting a right circular cylinder with a plane: geometric approach If you cut the cylinder x 2 + y 2 = 1 with a horizontal plane z = r, then the resulting curve of intersection is a circle with radius 1. But, if the plane is not quite horizontal, then the intersection is not a circle anymore, but looks more like an ellipse. But is it? (If the plane is vertical and meets the cylinder, the intersection consists of one or two lines; in the first case, the plane is tangent to the cylinder.) To decide on this matter, we first use a purely geometric approach, which uses the geometric definition of an ellipse: by a famous trick due to the French Belgian G.P. Dandelin ( ), we can locate two points which turn out to be the foci of the ellipse. Take two spheres of radius 1. Insert one in the cylinder from above, and the other from below. Move them down (resp. up) so that they touch the plane. Each sphere touches the plane in exactly one point, so we get two points, say F 1 and F 2. The spheres meet the cylinder in horizontal circles of radius 1. To show that the intersection of the cylinder and the plane is an ellipse, we take an arbitrary point P on the intersection and show that P F 1 + P F 2 is constant. Now, P F 1 equals the distance P B of P to the lower circle, and P F 2 equals the distance P A of P to the upper circle. But then P F 1 + P F 2 equals the distance between the two circles, which is obviously constant. In a similar way, all other types of intersections can be analysed. The right hand picture in Figure (3.17) illustrates the Dandelin approach to the intersection of a plane parallel to the line OC with a cone. The resulting intersection curve is a parabola. In the exercises we will discuss this case analytically Intersecting a right circular cylinder with a plane: analytic approach In terms of equations, things become complicated quite soon. To simplify matters, we rotate the whole situation so that the plane is given by the equation z = 0. For the sake of concreteness, let us assume that the cylinder has been rotated around the y axis over an angle of α radians. So if we rotate a point (u, v, w) on the rotated cylinder back over α radians the resulting coordinates should satisfy the equation x 2 + y 2 = 1. This means that

27 3.5 Geometric ins and outs on quadrics 83 O M 2 A F 2 D B A G F F 1 P M 1 B C H P Figure 3.17: On the left, Dandelin s approach to proving that the intersection of a plane with the right circular cylinder is an ellipse if the plane is not vertical. On the right, a similar illustration for a plane intersecting a cone along a parabola. This case is not explained in the text. (u cos α w sin α, v, u sin α + w cos α) satisfies x 2 + y 2 = 1: (u cos α w sin α) 2 + v 2 = 1. The interesection with the plane z = 0 is obtained by setting w = 0: evidently the equation of an ellipse. u 2 cos 2 α + v 2 = 1 and w = 0, Lines on hyperboloids Usually, the intersection of a plane with a quadric yields a truly curved curve. But sometimes, surprises occur. Let us illustrate this in the case of the hyperboloid of one sheet x 2 + y 2 z 2 = 1. Although this is a pretty curved object, it contains straight lines. For instance, the line (1, 0, 0) + λ(0, 1, 1) is on the hyperboloid. Just substitute (1, λ, λ) to verify this: λ 2 λ 2 = 1. This idea of lines on a hyperboloid was used in the design of the Vortex, which will possibly be built in England. Visually, the presence of lines on the hyperboloid may be clear,

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