﻿3. Mind Warm Ups - Conics

## Mind-Warm-Ups Conics: Recognizing Conics

Problem 1: what are conic sections or conics?
Ans: consider the double cone of figure 1. Some of the cross sections (plane slices) that can be obtained from the double cone are:
(a) Single point
(b) Single line
(c) Double lines
(d) Circle
(e) Parabola
(f) Ellipse
(g) Hyperbola
These shapes are called conic sections or conics. It is shown in the ellipse problems that the circle is a special case of an ellipse. The parabola, ellipse and hyperbola are called non-degenerate conics while the single point, single line and double lines are called degenerate conics

Problem 2: the standard forms of the equations of non-degenerate conics are different because the values of their respective eccentricities are different. What is meant by the eccentricity of a non-degenerate conic?
Ans: The eccentricity of a non-degenerate conic is a constant (denoted by e) that is incorporated into its definition. In general, non-degenerate conics are defined as the set of points P in R2 that satisfy the following distance equality:
PF = ePD -------(1).
Where PF is the distance from P to a fixed point F called the focus of the conic; PD is the distance from P to a fixed line called the directrix and e (the eccentricity) is a constant multiple of PD.
A non-degenerate conic is an ellipse if 0 ≤ e < 1 (a circle if e = 0), a parabola if e = 1, a hyperbola if e > 1.

Problem 3: Write a general equation that can represent the equations of non-degenerate conics.
Ans: Ax2 + Bxy + Cy2 + Fx + Gy + H = 0------(2). Where A, B and C are real numbers and are not all zero.

Problem 4 : given the general equation of conics:
Ax2 + Bxy + Cy2 + Fx + Gy + H = 0
and the associated matrices of figure 2.
(a) Write the general equation of conics in terms of the matrices A, J and x.
(b) How is the determinant of matrix A used to identify a conic?
Ans: (a) xTAx + JTx + H = 0.
(b) If detA < 0, conic is a hyperbola.
If detA = 0, conic is a parabola.
If detA > 0, conic is an ellipse.
Another way of stating the determinant test for identifying conics is as follows:
If B2 - 4AC > 0, conic is a hyperbola.
If B2 - 4AC = 0, conic is a parabola.
If B2 - 4AC < 0, conic is an ellipse.

Problem 5 : identify the following conics:
(a) 9x2 - 24xy + 6y2 - 6x + 12y - 48 = 0.
(b) 2x2 - 16xy + 32y2 - 2x + 16y - 24 = 0.
(c) 13x2 - 18xy + 19y2 - 8x + 18y - 28 = 0.
Ans:(a) A = 9, B = -24, C = 6
Therefore, B2 - 4AC = 576 - 216 = 360 > 0. Therefore conic is a hyperbola.
(b) A = 2, B = -16, C = 32.
Therefore, B2 - 4AC = 256 - 256 = 0. Therefore conic is a parabola.
(c) A = 13, B = -18, C = 19
Therefore, B2 - 4AC = 324 - 988 = -664 < 0. Therefore conic is an ellipse.

Problem 6: (a) can partial differential equations (PDEs) be expressed as conics?
(b) Write the general form for partial differential equations that are conics.
(c) What constraint is placed on conic partial differential equations?
Ans: (a) Yes.
(b) Auxx + Buxy + Cuyy + Dux + Euy + Fu = G
(c) The partial differential equations must be linear PDEs. That is, the dependent variable u and all its derivatives appear in a linear form in the eqaution (they are not multiplied together or squared).

Problem 7: what type of conic is the following PDE?
ut = uxx
Ans: A = 1, B = 0, C = 0. Therefore, B2 - 4AC = 0. Therefore, PDE is parabolic.

Problem 8: what type of conic is the following PDE?
uxx + uyy = 0.
Ans: A = 1, C = 1, B = 0.
Therefore, B2 - 4AC = -4 < 0. Therefore PDE is elliptic.

Problem 9: what type of conic is the following PDE?
uxy = 0
Ans: A = 0, B = 1, C = 0.
Therefore, B2 - 4AC = 1 > 0. Therefore PDE is hyperbolic.

Problem 10: what type of conic is the following PDE
yuxx + uyy = 0
Ans: No help here :).

Peter Oye Sagay