Matrix Algebra - Conic and Quadric Surfaces Determination, Study notes of Linear Algebra

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Matrix Algebra - 1016 - 331 - Fall 2005 Instructor: David S. Hart Geometry Handout We’ve seen that various curves and surfaces can be given by a determinant equation. For example, the line through €

( 1 ,^2 ) and

€

( 3,^4 ) is^ given^ by

€

x y 1 1 2 1 3 4 1 = 0. What follows is a summary of what we’ve done, along with further items you might want to explore. A conic curve in the plane has an equation of the form € a x 2

• bx y + c y 2
• d x + e y + f = 0 The expression € b 2 − 4 a c tells us what type of conic it is. In particular, three outcomes are given by € b 2 − 4 ac

0 hyperbola = 0 parabola < 0 ellipse











This must be used with care, and only after first making sure that we actually have a conic, and not something else. The equation € x + y = 1 satisfies € b 2 − 4 ac = 0 , yet it represents a line, not a parabola. A similar remark applies to € x 2

• 2 x y + y 2 = 1. Do you see what figure satisfies this equation? It is not a parabola, even though € b 2 − 4 ac = 0. Lastly, notice that € x 2
• y 2 = − 1 has no solutions in the real numbers. The fact that € b 2 − 4 ac = − 4 < 0 means nothing, and the equation certainly does not represent an ellipse. Identify the curves:

€ x 2

• xy + y 2 − 3 x = 4

€ x 2

• 2 xy + y 2 − 3 x = 4 - € x 2
• 3 xy + y 2 − 3 x = 4

In class I showed that it takes five points to determine a conic. Unless the conic has special properties, fewer points are not enough. Confirm this for yourself with the three curves just given. Find four points shared by all of them. So for each of these curves a fifth point on it will be the tie breaker!

• Pick five points, all with integer coordinates, on each of the above curves and write down the determinant form of the equation. Choose your points carefully, making good use of zeros. Evaluate the determinants you get.
• Remember that we can trade information for points. If we know something about a curve, then we need fewer points to determine its equation. A parabola that opens up or down has its central axis parallel to the y-axis, and its directrix parallel to the x-axis. Perhaps you recall some features of the equation of this type of parabola. If so, you will see at once that € b = 0 and € c = 0. Therefore, just three points are needed to find its equation. Write down the determinant form of the equation of such a parabola that passes through the points €

( −1,^10 ) ,

€

( 1 ,^4 ),^ and

€

( 3,^14 ).^ Evaluate^ the

determinant to get the usual form of the equation. The equation of a quadric surface in space has the form € a x 2

• b xy + c xz + dy 2
• e yz + f z 2
• g x + hy + iz + j = 0 So nine points, correctly chosen, will determine a quadric surface. The choice of points must not be made without care. Two different spheres, for example, may have infinitely many points in common. If so, then no number of points chosen from this common stock will distinguish them. Still, when wisely chosen, nine points suffice. Just as before, knowing what restrictions we have lessens the number of points we need. If it is a sphere we want, then € a = d = f and € b = c = e = 0. So a sphere can be determined by four points. Do you see why the four points must not all lie on a circle? If they do, then four points will not be enough to determine a sphere. In fact, no number of such points, all on a circle, would be enough.

• Consider the two surfaces € x 2

• y 2
• 2 z 2 = 25 , an ellipsoid € x 2
• y 2 − 2 z 2 = 25 , a hyperboloid of one sheet Find eight points, all with integer coordinates, that are shared by these two surfaces. On each surface find a ninth point, again with integer coordinates, that does not lie on the other. Use these points, or some modification of them, to give the determinant equation of each surface. Evaluate the determinants. Are you surprised by the results? In this problem, you may need several attempts to get the equations you expect. Try varying your selections of nine points in a thoughtful way. You may want to use a computer algebra system, like Maple, to help evaluate the determinants you get.

• If you enjoy problems like these, here is a last one. Use a determinant approach to find a simple equation of the quadric surface that passes through the following nine points. Identify the surface. € x 0 0 1 1 − 1 − 1 2 2 − 2 y 1 − 1 0 2 0 2 1 − 1 1 z − 1 − 1 0 − 6 2 − 4 − 1 − 1 3