Notes for Geometry Conic Sections The notes is taken from Geometry, by David A. Brannan, Matthew F. Esplen and Jeremy J. Gray, 2nd edition 1 Conic Sections A conic section is defined as the curve of intersection of a double cone with a plane. Figure: Examples: 1. non-degenerate conic sections: parabolas, ellipses or hyperbolas; 2. degenerate conic sections: the single point, single line and pair of lines. 1.1 Focus-Directrix Definition of the Non-Degenerate Conics The 3 non-degenerate conics can be defined as the set of points P in the plane satisfying: The distance of P from a fixed point F (called the focus of the conic) is a constant multiple (called its eccentricity, e) of the distance of P from a fixed line d (called its directrix). Eccentricity: A non-degenerate conic is 1. an ellipse if 0 ≤ e < 1, 2. a parabola if e = 1, 3. a hyperbola if e > 1. 1 Parabola (e = 1): A parabola is defined to be the set of points P in the plane whose distance from a fixed point F is equal to their distance from a fixed line d. We now derive a parabola in standard form: Let F (a; 0) be the focus and d : x = −a be the directrix. Let P (x; y) be an arbitrary point on the parabola and let M(−a; y) be the foot of the perpendicular from P to the directrix. Figure: Since FP = PM, by the definition of the parabola, this follows that FP 2 = PM 2 ) (x − a)2 + y2 = (x + a)2 ) y2 = 4ax: The point (at2; 2at); t 2 R lies on the parabola. 2 2 * (2at) = 4a · at : Conversely, we can write the coordinates of each point on the parabola in the form 2 y y2 (2at)2 2 (at ; 2at). For if we choose t = 2a , then y = 2at and x = 4a = 4a = at : It follows that there is a one-to-one correspondence between the real numbers t and the points of the parabola. Parabola in standard form A parabola in standard form has equation y2 = 4ax; where a > 0: It has focus (a; 0) and directrix x = −a; and it can be described by the parametric equations: 2 x = at ; y = 2at (t 2 R): We call the x-axis the axis of the parabola in standard form, since the parabola is symmetric with respect to this line. We call the origin the vertex of a parabola in standard form, since it is the point of the intersection of the axis of the parabola with the parabola. A parabola has no center. 2 Example 1.1. Write down the focus, vertex, axis and directrix of the parabola E with equation y2 = 2x: 1 1 Solution. Focus: F = ( 2 ; 0), Axis: x-axis, Vertex: (0; 0), Directrix: x = − 2 . Ellipse (0 ≤ e < 1): We define an ellipse with eccentricity zero to be a circle. We define an ellipse with eccentricity e (where 0 < e < 1) to be the set of points P in the plane whose distance from a fixed point F is e times their distance from a fixed line. We now derive an ellipse in standard form: a Let F (ae; 0); a > 0 be the focus and d : x = e . Let P (x; y) be an arbitrary point on a the parabola and let M( e ; y) be the foot of the perpendicular from P to the directrix. Figure: Since FP = e · PM, by the definition of the ellipse, this follows that a x2 y2 FP 2 = e2 · PM 2 ) (x − ae)2 + y2 = e2(x − )2 ) + = 1: e a2 a2(1 − e2) p Let b = a 1 − e2. Then x2 y2 + = 1: a2 b2 The equation is symmetrical in x and y. The ellipse also has a second focus F 0(−ae; 0) 0 a and a second directrix d : x = − e : We call the segment joining the points (±a; 0) the major axis of the ellipse and the segment joining the points (0; ±b) the minor axis of the ellipse. * b < a, the minor axis is shorter than the major axis. The origin is the center of this ellipse. Note that (a cos t; b sin t) lies on the ellipse. (a cos t)2 (b sin t)2 + = cos2 t + sin2 t = 1: * a2 b2 3 We can check that x = a cos t; y = b sin t; t 2 (−π; π] gives a parametric representation of the ellipse. Ellipse in standard form An ellipse in standard form has equation x2 y2 + = 1; where a ≥ b > 0; b2 = a2(1 − e2); 0 ≤ e < 1: a2 b2 It can be described by the parametric equations x = a cos t; y = −b sin t; t 2 (−π; π]: a If e > 0, it has foci (±ae; 0) and directrix x = ± e : Exercise 1.2. p(x + ae)2 + y2 + p(x − ae)2 + y2 = 2a: Hyperbola (e > 1): A hyperbola is the set of points P in the plane whose distance from a fixed point F is e times their distance from a fixed line d, where e > 1: We now derive a hyperbola in standard form: a Let F (ae; 0); a > 0 be the focus and d : x = e . Let P (x; y) be an arbitrary point on a the parabola and let M( e ; y) be the foot of the perpendicular from P to the directrix. Figure: Since FP = e · PM, by the definition of the hyperbola, this follows that a x2 y2 FP 2 = e2 · PM 2 ) (x − ae)2 + y2 = e2(x − )2 ) − = 1: e a2 a2(e2 − 1) p Let b = a e2 − 1. Then x2 y2 − = 1: a2 b2 4 The equation is symmetrical in x and y. The ellipse also has a second focus F 0(−ae; 0) 0 a and a second directrix d : x = − e : We call the segment joining the points (±a; 0) the major (transverse) axis of the hyperbola and the segment joining the points (0; ±b) the minor (conjugate) axis of the hyperbola. The origin is the center of this hyperbola. π π π 3π We can check that x = sec t; y = b tan t; t 2 (− 2 ; 2 ) [ ( 2 ; 2 ) gives a parametric representation of the hyperbola. The hyperbola consist of 2 branches. When x ! b ±∞, the branches get closer and closer to the lines y = ± a x the asymptotes of the hyperbola. Hyperbola in standard form A hyperbola in standard form has equation x2 y2 − = 1; where a ≥ b > 0; b2 = a2(e2 − 1); e > 1: a2 b2 a it has foci (±ae; 0) and directrix x = ± e : It can be described by the parametric equations π π π 3π x = a sec t; y = −b tan t; t 2 (− ; ) [ ( ; ): 2 2 2 2 Example 1.3. Determine the foci F and F 0 of the hyperbola E with equation x2 − 2y2 = 1: Solution. ±p3=2; 0 : Exercise 1.4. p(x + ae)2 + y2 − p(x − ae)2 + y2 = ±2a: 5.