Conic Sections Session 2: Ellipse

Conic Sections Session 2: Ellipse Toh Pee Choon NIE Oct 2017 Toh Pee Choon (NIE) Session 2: Ellipse Oct 2017 1 / 24

Introduction Problem 2.1 Let A, F 1 and F 2 be three points that form a triangle F 2 F 1 A. Find all points P such that the triangle F 2 F 1 P has the same area as F 2 F 1 A. Can you describe the locus of P? The locus of P is l: the line through A and parallel to the line F 2 F 1. (Is that all?) l the reflection of l about F 2 F 1 should also be included. Do we have a more elegant way to describe the locus? Define d as the line F 2 F 1 and for any point X recall that Xd denotes the perpendicular distance from X to d. The required locus is {P : Pd = Ad = k} for some real constant k. Toh Pee Choon (NIE) Session 2: Ellipse Oct 2017 2 / 24

Introduction Problem 2.2 Let A, F 1 and F 2 be three points that form a triangle F 2 F 1 A. Find all points P such that the triangle F 2 F 1 P has the same perimeter as F 2 F 1 A. Can you describe the locus of P? The locus of P is {P : PF 2 + PF 1 = AF 2 + AF 1 = k} for some real constant k. Can you describe the locus in geometric terms? It is an ellipse (passing through A) with F 2 and F 1 as its foci. Toh Pee Choon (NIE) Session 2: Ellipse Oct 2017 3 / 24

Definition of Ellipse Definition 2.3 (Geometric) An ellipse is the locus of points whose sum of distances to the foci F 1 and F 2 is a constant, i.e. {P : PF 1 + PF 2 = k}. B P A F 2 F 1 Toh Pee Choon (NIE) Session 2: Ellipse Oct 2017 4 / 24

Definition 2.4 An ellipse in standard position is defined by x 2 a 2 + y 2 = 1, where a b > 0. b2 The major and minor axes are segments through the center that correspond respectively to the largest and smallest distance between antipodal points on the ellipse. The vertices are the points of intersection of the ellipse and the major/minor axis. i.e. vertices are A = (a, 0), B = ( a, 0), C = (0, b) and D = (0, b). AB is the major axis and CD is the minor axis. (The analogous definitions hold when b > a.) C B (0, 0) A D Toh Pee Choon (NIE) Session 2: Ellipse Oct 2017 5 / 24

Problem 2.5 (Relating the Cartesian and Geometric definitions) An ellipse with major and minor axes on the x and y axes respectively and vertices (a, 0) and (0, b) is defined by PF 1 + PF 2 = k. Find the value of k and coordinates of F 1 and F 2. Hence calculate the focal length F 1 F 2. The coordinates of F 1 and F 2 must be (c, 0) and ( c, 0) respectively. When P = (a, 0), AF 1 + AF 2 = a c + (a + c) = 2a. When P = (0, b), by Pythagoras Theorem, BF 1 + BF 2 = 2 c 2 + b 2 = 2a. Hence k = 2a and c = a 2 b 2. Finally the focal length is 2 a 2 b 2. Toh Pee Choon (NIE) Session 2: Ellipse Oct 2017 6 / 24

Problem 2.6 (Geometric = Cartesian) Derive the Cartesian equation from the Geometric definition. Since c = a 2 b 2, PF 1 + PF 2 = 2a, we have (x c) 2 + y 2 + (x + c) 2 + y 2 = 2a = (x + c) 2 + y 2 = (2a ) 2 (x c) 2 + y 2 = 4a 2 4a (x c) 2 + y 2 + (x c) 2 + y 2 = 4xc 4a 2 = 4a (x c) 2 + y 2 = ( a 2 cx ) 2 = a 2 (x c) 2 + a 2 y 2 = a 4 + c 2 x 2 = a 2 x 2 + a 2 c 2 + a 2 y 2 = a 2 (a 2 c 2 ) = (a 2 c 2 )x 2 + a 2 y 2 = a 2 b 2 = b 2 x 2 + a 2 y 2. Toh Pee Choon (NIE) Session 2: Ellipse Oct 2017 7 / 24

Exercise 2.7 Let x 2 a 2 + y 2 b 2 = 1, a b > 0 represent an ellipse in standard position and F be a focus. Find the length of the chord parallel to minor axis and passing through F. This chord is called a latus rectum. Consider F 1 = (c, 0) where c = a 2 b 2 So y 2 = 1 a2 b 2 b 2 a 2 i.e. y = ± b2 a. So the latus rectum is of length 2b2 a. Toh Pee Choon (NIE) Session 2: Ellipse Oct 2017 8 / 24

Exercise 2.8 If the major axis of an ellipse given by the equation x 2 4 + t + y 2 t 2 2 = 1 lies on the y-axis, find the range of values of t. We require 4 + t > 0 = t > 4 and t 2 2 > 0 = t < 2 or t > 2. Furthermore if the major axis is on the y-axis, we need t 2 2 4 + t. i.e. t 2 t 6 0 = t 3 or t 2. Combining the two gives t 3 or 4 < t 2. Remark: In the special case of a circle (t = 3 and 2), every possible diameter can be considered a major axis. We use GeoGebra to verify our calculations. Toh Pee Choon (NIE) Session 2: Ellipse Oct 2017 9 / 24

Exercise 2.9 Use Geogebra to plot an ellipse with foci F 1 = (5, 0) and F 2 = (3, 0) and major axis of length 4. Consider an arbitrary point P on this ellipse and join the segment PF 2. By varying the position of P, deduce a relation between PF 2 and the coordinates of P. Since the foci must lie on the major axis, one of the vertex of this ellipse must lie on (6, 0). It appears that PF 2 is always half the value of the x-coordinate of P. In other words, any point P is always twice as far away from the y-axis as its distance to F 2. Define d as the line x = 0 and denote Pd as the perpendicular distance of P to d. Then Pd = 2 PF 2. To see this clearly, define D = (0, y(p)), i.e. D is on the directrix and has the y-coordinate as P. Toh Pee Choon (NIE) Session 2: Ellipse Oct 2017 10 / 24

Exercise 2.10 Let F = (0, 0) and d be the line x = 4. Find the locus of points whose distance from d is thrice the distance from F. i.e. {P : Pd = 3 PF }. (1, 0) satisfy the equation and so does ( 2, 0). It s reasonable to assume the locus is continuous. From (1, 0), where would the neighbouring points lie? Let P = (0, y) and Pd = 4, so we have y = ± 4 3. These four points are insufficient to determine a conic. (Use GeoGebra tool to see this.) For general P = (x, y), we have 3 ( x 2 + y 2 = 4 x = 8 x + 2) 1 2 + 9y 2 = 18. Toh Pee Choon (NIE) Session 2: Ellipse Oct 2017 11 / 24

Definition of Ellipse Definition 2.11 (Directrix-Eccentricity-Focus) An ellipse is the locus of points whose distance from a fixed point F (focus) and to a fixed line d (directrix) are in a constant ratio less than 1. This ratio is termed the eccentricity. i.e. { P : PF } Pd = e, 0 < e < 1. d P D F Toh Pee Choon (NIE) Session 2: Ellipse Oct 2017 12 / 24

Linking the definitions Problem 2.12 (D-E-F = Cartesian) Let F = (c, 0) and directrix d be the line x = k. Transform the equation PF Pd = e for e 1 to the form (x c ) 2 ke2 1 e 2 + y 2 1 e 2 = e2 (k c) 2 (1 e 2 ) 2. Let P = (x, y) then (x c) 2 + y 2 = e 2 (k x) 2. (Note the consequence of e = 0. Alternatively, set e = 1 k and let k.) This becomes x 2 (1 e 2 ) 2x(c ke 2 ) + y 2 = e 2 k 2 c 2. (Note the consequence of e = 1.) When 0 < e < 1, this is an ellipse thus unifying both definitions. For standard position, we set k = c/e 2. The equation simplifies to e 2 x 2 e 2 y 2 c 2 + c 2 (1 e 2 ) = 1. Toh Pee Choon (NIE) Session 2: Ellipse Oct 2017 13 / 24

Linking the definitions Definition 2.13 (D-E-F = Cartesian) Let F = (c, 0), c > 0 and directrix d be the line x = c e 2. The equation PF Pd = e for 0 < e < 1 is equivalent to e 2 x 2 e 2 y 2 c 2 + c 2 (1 e 2 ) = 1 Thus a = c e and b = (1 e 2 )a. (i.e. a > b > 0.) Directrix becomes x = a e. e = 1 b2 a 2 c = ae = a 2 b 2. (So the focus defined in the G and D-E-F definitions coincide.) By symmetry F 2 = ( c, 0) and x = a e also qualify as focus and directrix. Toh Pee Choon (NIE) Session 2: Ellipse Oct 2017 14 / 24

Linking the definitions Problem 2.14 (Cartesian = D-E-F ) Given an ellipse in standard position i.e. x2 + y 2 = 1, a b > 0 show for a 2 b 2 c = a 2 b 2 and there exist some e, satisfying 0 < e < 1 such that (x c) 2 + y 2 = e 2 ( c e 2 x ) 2. Define e = c a, which satisfies the bounds given. This gives b = a 2 c 2 = a (1 e 2 ) = c e (1 e 2 ). One can then work backwards to obtain the desired equation. In conclusion the D-E-F definition holds with F = (c, 0) and directrix x = c e 2. Toh Pee Choon (NIE) Session 2: Ellipse Oct 2017 15 / 24

PF = e Pd : Varying eccentricity and directrix Exercise 2.15 In GeoGebra, create sliders with variables 0 e 5 and 5 k 20. Create the directrix x = k and focus F = (0, 0). Create a slider 0 m 100, where m is the value of Pd. This means that P lies on one of the lines x = k + m or x = k m. (Create these two lines.) On the other hand, P lies on a circle of radius em with centre F. The intersections (four possibilities) of this circle with the previous two lines gives P. Vary m and use trace to visualize the conic. Toh Pee Choon (NIE) Session 2: Ellipse Oct 2017 16 / 24

Definition of Ellipse Definition 2.16 (Scaled Circle) Given a circle and a straight line through the centre. If we apply a scaling of factor k in the direction of the line, we obtain an ellipse. Scaling of factor b a parallel to the y-axis of the circle x 2 + y 2 = a 2. Q b P O t a P = (a cos t, b sin t) Toh Pee Choon (NIE) Session 2: Ellipse Oct 2017 17 / 24

Parametric Form Definition 2.17 In the diagram, if Q = (a cos t, a sin t) for π < t π, then the coordinates of P is P = (a cos t, b sin t). Important: t is not equal to θ, the angle between OP and the x-axis. Q b P O t θ a tan θ = b a tan t Toh Pee Choon (NIE) Session 2: Ellipse Oct 2017 18 / 24

Linking the definitions Problem 2.18 (Cartesian = Geometric ) Let x2 = 1 be an ellipse with a b > 0 and let F b 2 1 be the right focus. Show that for an arbitrary point P = (a cos t, b sin t), a 2 + y 2 PF 1 = a(1 e cos t). PF 1 = (a cos t a 2 b 2 ) 2 + b 2 sin 2 t = a 2 cos 2 t 2a(cos t) a 2 b 2 + a 2 b 2 + b 2 (1 cos 2 t) ( a = 2 b 2 cos t a ( 2 = a 1 b2 cos t 1) = a(1 e cos t). a 2 ) 2 Similarly PF 2 = a(1 + e cos t). Consequently, PF 1 + PF 2 = 2a. Toh Pee Choon (NIE) Session 2: Ellipse Oct 2017 19 / 24

Reflection Property Problem 2.19 Given an ellipse in standard position with foci F 1 and F 2. Let P = (x 0, y 0 ) be a point on the ellipse and N be on the x-axis such that PN is normal to the tangent line of the ellipse at P. Show that F 2 PN = NPF 1. P N F 2 F1 Aim: Show PF 2 NF 2 = PF 1 NF 1. Toh Pee Choon (NIE) Session 2: Ellipse Oct 2017 20 / 24

Proof of Reflection Property Theorem 2.20 (Angle Bisector Theorem) In ABC with D on BC,AD bisects BAC if and only if AB BD = AC CD. Aim: Show PF 2 NF 2 = PF 1 NF 1. We have PF 1 = a ex 0 and PF 2 = a + ex 0. Differentiating implicitly, we have 2 x a 2 + 2 y b 2 dy dx = 0 = dy dx = xb2 ya 2. So the gradient of PN is given by a2 y 0 b 2 x 0. Let N = (x 1, 0), then ) y 0 = a2 y 0 x 0 x 1 b 2 = x 1 = x 0 (1 b2 x 0 a 2 = e 2 x 0. NF 1 = c x 1 = ae e 2 x 0 = e(a ex 0 ) and NF 2 = x 1 + c = e 2 x 0 + ea = e(a + ex 0 ). Toh Pee Choon (NIE) Session 2: Ellipse Oct 2017 21 / 24

Polar form with respect to the focus Problem 2.21 Let F 1 = (0, 0) be a focus and x = k, k > 0 be the directrix. Find the polar form of the resulting ellipse with eccentricity e. r P F2 F1 θ k PF 1 = r and P d = k r cos θ. So r = e(k r cos θ) = r = ek 1 + e cos θ. Toh Pee Choon (NIE) Session 2: Ellipse Oct 2017 22 / 24

Problem 2.22 In the diagram AB is the major axis of an ellipse in standard position. The tangent to the ellipse at P intersects the tangents to vertices A and B at C and D respectively. Show that PF 1 PF 2 = PC PD. C P D A F 2 F 1 B Toh Pee Choon (NIE) Session 2: Ellipse Oct 2017 23 / 24

Problem 2.23 Cut out a paper disc and mark out an arbitrary interior point X distinct from the centre of the disc. Now pick a random point P on the circumference of the disc and and fold the disc (with one crease) such that P touches X. Use a pencil to mark out the crease. Repeat for as many points P as possible. What do you observe? Explain. Toh Pee Choon (NIE) Session 2: Ellipse Oct 2017 24 / 24