Samantha Klein 2009 The Conic Sections
Have you ever heard someone say, “ I was great at algebra but terrible at geometry” or visa versa? In mathematics there are topics taught in a geometry class or an algebra class. The conics sections are one of the topics in mathematics that can be taught in either a geometry or algebra class. Each of the conics has both a geometric definitions and an algebraic definition. In this paper we will look at the definitions of each conic section and proofs of how the definitions are related to each other. Geometric definitions: A conic section is a curve created by the intersection of a plane and a right conical surface. There are four curves that can be created when intersecting a plane with a cone (conical surface), a circle, ellipse, parabola and hyperbola. Some argue that the circle is a distinct fourth conic section, other argue that the circle is a special case of the ellipse; for this paper the circle will be considered a special case of the ellipse. Changing the angle of the intersecting plane with the right cone creates the conic sections. Let’s look at this definition in more detail. First start by creating a cone. To create a cone you need to start with a vertical line called an axis. Then a second line called a generator will intersect the axis at some angle, called the vertex angle or . Let the generator be attached to the axis such that if you were to grab the top of the axis and spin it around 360o, the generator will also rotate with the axis, thus creating a conical surface or cone with two nappes and a vertex where the two nappes meet.
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GENERATOR
Once the cone is formed we can then cut it with a flat plane at certain angles to create the conic sections. A plane that is parallel to the generator (side of the cone) will produce an open curve called a parabola. If the plane cuts through the cone at an angle with the axis that is between perpendicular to the axis and parallel to the generator then the result is a closed curve known as an ellipse. The special case of the ellipse, the circle, is formed when the plane’s angle is perpendicular to the axis. In all the above cases the plane will only cut through one nappe of the cone. The last conic, the hyperbola, is the one conic that is produced by cutting through both nappes of a cone. In order to cut through both nappes of the cone the angle between the plane and the axis must be smaller than the vertex angle. (See Figure 1)
- 1 - Samantha Klein Figure 1 2009
Parabola Ellipse Circle Hyperbola Angle: Angle: Special case of Angle: Parallel to Between ellipse Smaller than generator (side Perpendicular to Angle: vertex angle of cone) axis and parallel Perpendicular to generator to axis
The parabola, ellipse, circle and hyperbola are all curves that are formed by planes cutting through the nappes of a cone. These planes never go through the vertex of the cone. If the plane does go through the vertex of a cone then we create what are called the degenerate conics. The degenerate conics consist of a point, line and two intersecting lines. If the plane is perpendicular to the axis and goes through the vertex we will create a point. If the plane is parallel to the side of the cone and goes through the vertex we will get a line. Last if the plane is not parallel to the side or perpendicular to the axis and goes through the vertex then the resulting degenerate conic is two intersecting lines. (See Figure 2) Figure 2
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Point Line Two
intersecting lines
According to “The History of Greek mathematics, Volume II “ by Sir Thomas Little Heath the discovery of the conics is said to have been by a Greek mathematician Menaechmus in about 360 or 350 B.C. (page 116). He discovered the conics in his attempts to solve one of the three problems of antiquity, doubling the cube (page 110). The construction of the conics that Menaechmus studied were a little different than what
- 2 - Samantha Klein 2009 we think of as the conics today. In order for Menaechmus to get a desired conic section he would leave the plane at a right angle to one of the generators and then change the angle of the cone, where as today we change the angle of the plane and use a right cone to create the conic sections. Apollonius of Perga known as “The Great Geometer” published his great work Conics in eight volumes. The first four books of the Conics is said to be a compilation of all the work done before him. Although according to Apollonius himself the books are “... worked out more fully and generally than in the writings of others.” It is in this work Conics that Apollonius introduces the terms we use today of parabola, ellipse and hyperbola. He was also credited with changing the idea that conic sections could be formed by changing the angle of the plane as opposed to changing the angle of the cone. That is he states what we use today as a geometric definition of the conic sections. Among other famous Greek mathematician to study the conic sections were Aristaeus and Euclid. Euclid made a compilation or rearrangement of all the work on conics known to him in his time (Heath, 116). According to “Selections Illustrating the History of Greek mathematics, Volume II” by Ivor Bulmer-Thomas, “Aristaeus was obviously more original and more specialized; that of Euclid was admittedly a compilation largely based on Aristaeus” (page 281). According the Pappus the first four books of Apollonius’s Conics were based on the first four books of Euclid’s work (Heath, 119). Interestingly, it is noted by Pappus that the focus – directrix property must have been known to Euclid, and probably to Aristaeus, but is never mentioned in Apollonius’s Conics. The focus- directrix property is what we think of today as the “plane geometry” or locus definition of the conics. “ Pappus’s enunciation of the theorem is to the effect that the locus of a point such that its distance from a fixed point is a given ratio to its distance from a fixed straight line is a conic” (Heath, 119). Today we refer to this as the property of eccentricity. Before we look at the idea of eccentricity let’s first look at each individual conic and its “plane geometry” definition. In plane geometry the ellipse is defined as the locus of a point on a plane, in which the sum of the distance from two fixed points, called foci, remains a constant 2a. The circle, the special case, is defined as the locus of a point on a plane that is a fixed distance from a fixed point (its center). A hyperbola is defined as the locus of a point in a plane where the difference in the distance from any point to two fixed points, called foci, remains a constant 2a. The parabola is defined as the locus of a point on a plane equidistant from a fixed point (focus) and a fixed line (directrix). Although these locus definitions seem to have no connection to the definitions in terms of the intersection of a plane and a cone that we mentioned earlier. We know that Menaechmus knew that the conic sections were formed by a plane intersecting a cone, and according to Pappus, Euclid must have know the focus- directrix property (locus definition). Therefore one can only assume that the connection between these two definitions was known prior to the 19th century. However, it was not until 1822 when the Belgian mathematician Germinal Pierre Dandelin was accredited with finding a simple proof using tangents and spheres in a cone to prove the connections between the locus definitions and the intersections of a plane and cone for the conics. Before we can do this proof we must get some background information out of the way. We must first show the proof of why the length of two tangents drawn from an exterior point to a circle or sphere are equal. This is a key idea used by Dandelin in his proof.
- 3 - Samantha Klein 2009 Proof that the length of two tangents drawn from an exterior point to a circle or sphere are equal. Given that AB and AC are tangent to circle O at points from common exterior point A. Figure 3 Then we can prove that AB = AC. First construct segment AO. Then construct BO and CO, which by definition are both radii r of circle O, so OB r OC (See Figure 3). Since a tangent is defined as a being r perpendicular to the radius then AB OB and AC OC. Now we have two right triangles ABO and ACO, which share a side AO. Since ABO and ACO are 2 2 right triangles we can use Pythagorean theorem to show that AB r 2 AO so 2 2 2 2 AB AO r 2 and AC r 2 AO so AC AO r 2 . Since 2 2 AB AO r 2 and AC AO r 2 , then AB = AC by the transitive property of equality. Hence, The lengths of the two tangents drawn from a common exterior point to a circle are equal. Q.E.D. The same proof can be used in three dimensions to show that the tangents from an exterior point to a point on a sphere are equal. Given two lines through a common point, we can construct a plane that contains these two lines and the point. Now, set the lines tangent to a sphere. The plane will intersect the sphere creating a circle. Because they are tangent to the sphere the lines must also be tangent to the circle thus the above theorem hold true in both two dimensions and three dimensions. Hence, the lengths of two tangents from an exterior point to a sphere are equal. Dandelin’s proofs will require us to use either one or two spheres (depending on the number of foci) placed inside of a cone. From above we can show that the tangents from an exterior point to either a smaller or larger sphere in the cone are equal or constant in length (See Figure 4 and 5). With some simple subtraction we can also show that the lengths from Figure 4 Figure 5 Figure 6 one sphere to the other are also equal or constant. If you were to take the lengths in figure 5 and subtract the lengths in figure 4 you will show that the lengths from the small sphere to the larger sphere are also equal or have a constant length QuickTime™ and a QuickTime™ and a QuickTime™ and a decompressor decompressor decompressor (See Figure 6). This can also be shown because are needed to see this picture. are needed to see this picture. are needed to see this picture. the two planes that are creating the circles are parallel. We know they are parallel because both planes must be perpendicular to the axis of the cone. Since the planes are parallel then the distance along the cone between the two circles must be a constant. Now we are ready to start looking at Dandelin’s Proofs. We will start with the ellipse.
- 4 - Samantha Klein 2009 Ellipse and Circle Dandelin’s Proof connecting two geometric definitions of an ellipse Let’s review the two definitions of the ellipse and what we are trying to prove. The section definition of and ellipse is as the intersection of a plane with a Figure 7 cone at an angle that is between perpendicular to the axis and parallel to the generator (See Figure 7). The locus definition is as the locus of a point on a plane,
QuickTime™ and a such that the sum of the distance from two fixed points, called foci, remains a decompressor constant 2a. Here 2a is the length of the major axis, a fact that we will prove little are needed to see this picture. later. Let’s start by proving the two definitions are related to each other. To do so we will need to intersect a cone with a plane at angle that is between perpendicular to the axis of the cone and parallel to the generator. Then we construct two spheres inside the cone, which are tangent to the plane of ellipse and tangent to the cone simultaneously. A smaller sphere will be placed above the ellipse, tangent to the plane of the ellipse at exactly one point, E, which is one of the two foci, and tangent to the cone where it forms the circle O. The second and larger sphere will be below the ellipse and tangent to the plane of the ellipse at the other focus D and tangent to the cone at circle P (See Figure QuickTime™ and a decompressor 8). Let point B be any point on the ellipse. are needed to see this picture. We must now show that DB EB a constant as point B move around the ellipse. Construct a line along the side of the cone from the vertex V, which intersects circle O, point B and circle P. We will call the point where the line interests with circle O, point A. The point where the line intersects circle P we will call, point C. Now, as point B moves around the ellipse, point A and C will be moving around circles O and P, respectively. Since AB and BE are both tangents to the same sphere from an exterior point then AB BE . For the same reason we can say that BD BC . We showed earlier that the lengths of VA and VC are constant no matter what line we chose on the side of the cone. Since VC VA AC then AC must also be constant. Now, we know that AC AB BC and since AB BE and BD BC , by substitution AC BE BD. We know AC is a constant, therefore BE BD is a constant. Hence, an ellipse is the locus of a point on a plane, in which the sum of the distance from two fixed points, called foci, remains a constant and is the intersection of a plane with a cone at an angle between perpendicular to the axis of the cone and parallel to the generator. Q.E.D.
- 5 - Samantha Klein 2009 Above we stated that an ellipse is the Figure 9 Point (x,y) locus of a point on a plane, in which the sum of the distance from two fixed points, called foci, d1 d 2 remains a constant 2a. In the proof above we Major Axis F2 simply showed that it was a constant, now we F1 need to prove that the constant is 2a, or the length of the major axis. To prove the fact that Minor Axis the sum of the two distances from a point on the ellipse to the foci remains a constant 2a, we need to start with some definitions. Every ellipse has what is called the major axis and minor axis. The major axis is the longest diameter of the ellipse and the minor axis is the shortest diameter of the ellipse (See figure 9). The major axis contains both the center and the two foci of the ellipse. The center of the Figure 10 ellipse is defined as the midpoint between the Point (x,y) two foci, or the point of intersection between
b d2 the line through the foci and its perpendicular d1 c bisector. Therefore the ellipse has two foci, F F 1 2 and F , one on either side of the center, F 2 1 equidistand from the center. Suppose the length a Center from the center of the ellipse to a focus is c. Let the length from the center to the point on the ellipse and major axis be a, and the length from the center to the point on the ellipse and minor axis be b (See figure 10). Then by the symmetry of the ellipse we know that the length of major axis is 2a.
Proof that the sum of the distance from a point on an ellipse to the two foci is a constant length of 2a. Now we will prove that the sum of the lengths from a point on the ellipse to the two foci is 2a, or the length of the major axis. In other words we need to show that d1 +d2 = 2a. Suppose we were to move the point so that it is located on the ellipse and major axis Figure 11 (See Figure 11). Since the point is on the major we know that d1 = c + a, and d2 = a – c. By substitution d1+d2 = (c + a) +(a – c). c d1 F2 Combining like terms yields d1+d2 = 2a. F1 d2 Hence, the sum of the distances from a point c a on an ellipse to the two foci is a constant 2a, the length of the major axis. Q.E.D.
Derivation of the standard algebraic form of the ellipse. From here we can also derive the standard algebraic form for the ellipse. All we need is to label some points. Lets start by putting the ellipse centered at the origin (0,0). Let the distance from the center to a focus be c. So the foci are located at (c, 0) and (-c, 0). Now, let the distance from the center to the points on the ellipse and the major axis be a, where a > c. We will call these points (-a, 0) and (a, 0). Let the distance from
- 6 - Samantha Klein 2009 the center to the points on the ellipse and the Figure 12 (0,b) minor axis be b where a > b, call these two Point (x,y)
points (0, -b) and (0, b) (See Figure 12). We d1 will call a point on the ellipse (x, y). d2 By placing the point (x, y) at the same (-a,0) (-c,0) (0,0) (c,0) (a,0) location as (0,b) (See Figure 13) we can now say that d1 d2 , and using the fact that (0,-b) d1 d2 2a, we can show that d1 a = d2. Using Pythagorean theorem we know that 2 2 2 b a c , which we will use later in the Figure 13 Point (x, y) proof. (0,b) Using the distance formula we know d1 = a d = a 2 2 2 2 b 2 d1 (x c) y and d2 (x c) y .
Substituting we can say if d1 d2 2a, then (-a,0) (-c,0) (0,0) c (c,0) (a, 0) x c 2 y 2 x c 2 y 2 2a.
(0,-b) Simplify and, x 2 2xc c 2 y 2 x 2 2xc c 2 y 2 2a. 2 2 Rearranging and squaring both sides, x 2 2xc c 2 y 2 2a x 2 2xc c 2 y 2 yields, x 2 2xc c 2 y 2 4a2 4ax2 2xc c 2 y 2 x 2 2xc c 2 y 2 . Simplifying and combining like terms leaves 4xc 4a2 4ax2 2xc c 2 y 2 . Divide both sides by four and move a2 to the opposite side, so now we have a2 xc ax2 2xc c 2 y 2 . Again we need to square both sides and simplify so that a2 x 2 x 2c 2 a2c 2 a2 y 2 a4 . Regrouping and factoring shows x 2a2 c 2 a2 y 2 a2a2 c 2. From earlier, we used Pythagorean theorem to show that b2 a2 c 2 , thus x 2a2 c 2 a2 y 2 a2a2 c 2 is the same as x 2b2 a2 y 2 a2b2. x 2 y 2 Dividing both sides by a2b2 and we are left with 1, which is the standard a2 b2 algebraic form of an ellipse centered at the origin.
Now suppose the center is located at some point (h, k) (See Figure 14), then using the same proof as above yield the Figure 14 Point (x, y) standard algebraic form of any ellipse (h, k+b) centered at (h, k). d1 = a d = a b 2 (x h)2 (y k)2 2 2 1 a b (h-a, k) (h-c, k) (h,k) c (h+c, k) (h+a, 0)
(h, k-b)
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The Circle: A special case of the ellipse. The special case of the ellipse, the circle, has a very similar proof to the one shown above. A circle is the locus of a point on a plane that is a fixed distance from a fixed point (its center), when the plane cuts through the cone at a perpendicular angle to the axis. Unlike the proof of the ellipse, in the proof of the circle the two spheres would touch the plane at the same point (the center of the circle) as opposed to two points the foci (See Figure 15). This is because the circle is a special case of an ellipse in which the two foci coincide, thus forming the center of a circle. You could use either sphere to show that the radius, the distance from the center to a point on the circle, is a constant length.
Derivation of standard algebraic form of the circle
Algebraically, a circle is formed when a = b in the standard for of the ellipse, causing the foci to coincide. By definition it is the locus of a point on a plane that is a fixed distance from a fixed point (its center). Using this definition we can derive the standard algebraic form of a circle. To start, construct a circle centered at (h, k). Let P be any Figure 16 point on the circle located at (x, y) (See Figure 16). By definition the distance from point P to the center is the y P (x, y) radius, r. The distance formula between two point on the coordinate plane tells us that r r x h 2 y k 2 . Square both sides and we x have derived the standard algebraic form for a circle (h, k) centered at (h, k), r2 x h 2 y k2 .
The Hyperbola Just as with the ellipse the hyperbola has both a section definition and a locus definition that can be proven related through Dandelin’s spheres. The hyperbola’s section definition is the intersection of a plane with a cone in which the angle of intersection is less than the vertex angle. The locus definition is as the locus of a point on a plane, such that the difference of the distance from two fixed points, called foci, remains a constant 2a. Here 2a is also the length of the transverse axis. This again will be proven later. For now we need to show the difference of the distance from two fixed points is a constant. The hyperbola is basically the opposite of the ellipse. The ellipse is the sum of the distances from a point to the two foci is a constant. The hyperbola on the other hand, is
- 8 - Samantha Klein 2009 the difference between the distances from a point to the two foci is a constant. The ellipse is also a closed curve whereas the hyperbola is an open curve. In Dandelin’s proof of the ellipse we used two spheres, one for each focus, but all in one nappe of the cone. The hyperbola will also need two spheres, one for each focus, however in the case of the hyperbola each sphere will be in opposite nappes of the cone (See Figure 17).
Dandelin’s Proof connecting the two geometric definitions of a hyperbola.
Figure 17 Figure 17 shows a plane intersecting with a cone that and angle that is less than the vertex angle. To show that the curved lines produced by this intersection is a hyperbola, we must show that the difference of the distance from two fixed points, called foci, remains a constant. We will begin by constructing two Circle M spheres, on in each nappe of the cone. The first
QuickTime™ and a sphere will be constructed in the top nappe of decompressor are needed to see this picture. the cone and will be tangent to the cone at the circle M and tangent to the intersecting plane at the foci, F1, simultaneously. The second sphere Circle N will be constructed in the bottom nappe of the cone and will be tangent to the cone at the circle N and tangent to the intersecting plane at the foci, F2, simultaneously. Let P be any point on the hyperbola, we must now show that
PF1 PF2 = a constant as point P moves about Copyright © 2009, Louis A Talman the hyperbola. First, construct a line along the side of the cone from point P that intersects the second sphere on circle N, the vertex V and with circle M on the first sphere. The point where the line intersect circle N we will call point B. The point where the line intersects circle M we will call point A. As point P moves along the hyperbola, points A and B will rotate around the circle M and N, respectively. Now, since PBand PF2 are both tangent
to the same sphere from a common exterior point then PB PF2 . For the same reason
we can say PA PF1 . Since we know PB PF2 and PA PF1 , then by substitution
PF1 - PF2 = PA PB = AB. Circle M and circle N are both tangent to the cone then they must be parallel, therefore the line segment AB on the side of the cone that connects them must remain a constant, thus AB is a constant. We know that AB is a constant,
therefore PF1 PF2 = a constant. Hence, a hyperbola is the locus of a point on a plane, in which the difference of the distances from two fixed points remain a constant and is the intersection of a plane and a cone when the angle of intersection is less than the vertex angle of the cone. Q.E.D.
The above proof shows that the locus definition of a hyperbola does have a connection to the intersection of a cone definition. We stated early that the difference between the distances from a point on the curve to the two foci is a constant, 2a. Now we
- 9 - Samantha Klein 2009 need to show that the constant is 2a, the length of the transverse axis. Just as the ellipse, the hyperbola has a longer and shorter axis, which are referred to as the major and minor axis, respectively. The transverse axis of the hyperbola is the segment of the major axis that connects the two vertices of the hyperbola. The vertices are the points where the hyperbola intersects with the major axis. For a hyperbola that opens to the left and right the transverse axis would be along the x-axis and for a hyperbola that opens up and down the transverse axis would be along the y-axis. The perpendicular bisector of the transverse axis will produce the conjugate axis and the center of the hyperbola. On either side of the center you will find a vertex and focus of the hyperbola (See figure 18).
Figure 18
Conjugate Axis
Vertex Vertex QuickTime™Center and a decompressor Major are needed to see this picture. Axis Transverse Axis
Minor Axis
Proof the distance from a point on a hyperbola to two foci is a constant 2a, the length of the transverse axis. Suppose the center of the hyperbola is at the origin, O, let the distance from the center to a vertex be a distance of a, therefore my vertices are at (-a,0) and (a,0). By the definition, both the foci and the vertices will be symmetrical about the center of the hyperbola. Now, let P be any point on the hyperbola. Place point P at the vertex, such that P is now at (a, 0) (see figure 19). By doing so d1 =c + a and d2 =c - a. By definition of the hyperbola, we need to show that d1 –d2 = 2a. By substitution d1 –d2 =c + a-(c - a) = c +a –c +a. Combing like terms yields, d1 –d2 = 2a, which in the case of the hyperbola is the length of the transverse axis. Hence, the difference of the distances from a point on a hyperbola to two fixed foci is a constant 2a, the length of the transverse axis. Q.E.D.
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Figure 19
P(x,y) now P(a,0) (0,b)
Vertex b (-c, 0) Vertex (c, 0) (-a,0) dQuickTime™1 and (a,0)a d decompressor 2 are needed to see this picture. a (0,-b)
Derivation of standard algebraic form of the hyperbola. Let the center of the hyperbola be placed at origin, O. Then the major axis be the x axis. Let the distance from the center to a foci be c. Therefore the foci will be located at (-c, 0) and (c, 0). Let the vertex be a distance of a from then center where a Figure 20 P (x,y) (0,b) d1 d2 Vertex (-c, 0) Vertex (c, 0) (-a,0) QuickTime™ and (a,0)a decompressor are needed to see this picture. (0,-b) Center 2 2 2 2 Using the distance formula d1 x (c) y 0 x c y and 2 2 2 2 d2 x c y 0 x c y . We showed early that d1 d2 2a, thus by substitution x c 2 y 2 x c 2 y 2 2a. Now to remove the radicals we will - 11 - Samantha Klein 2009 move one radical to the opposite side x c 2 y 2 2a x c 2 y 2 . Now square 2 2 2 2 both sides x c y 2 2a x c y 2 and you get x c 2 y 2 4a2 4a (x c)2 y 2 (x c)2 y 2 . Multiply out the squares x 2 2xc c 2 y 2 4a2 4ax c 2 y 2 x 2 2xc c 2 y 2. Now by isolating the variable, most terms will cancel out leaving 4xc 4a2 4ax c 2 y 2 . Now, divide everything by 4, xc a2 ax c 2 y 2 . We still need to get rid of the radical so we have to square both sides again, which leaves us x 2c 2 2a2 xc a4 a2x c 2 y 2. Simplify and then distribute the a2, so x 2c 2 2a2 xc a4 a2 x 2 2a2 xc a2c 2 a2 y 2 . Now, rearrange so that all the constants are on one side x 2c 2 a2 x 2 a2 y 2 a2c 2 a4 . Rearranging the variables was a clever way to allow us to factor out and x2 from the first two terms and an a2 from the left side x 2c 2 a2 a2 y 2 a2 c 2 a2 . Since we know that a Figure 21 Minor Axis (h,b+k) P (x,y) d1 d2 (-c+x,k) (c+h,k) (-a+x,k)QuickTime™ andV(a+h,k) a decompressor are needed Center to see ( hthis,k) picture. Major Axis (h,-b+k) - 12 - Samantha Klein 2009 The Parabola The last of the conic sections we will look at it is the parabola. A parabola is defined as the intersection of a cone with a plane that is parallel to its generator (side of the cone). The parabola has a locus definition that is different from the two previous conic sections discussed. The locus definition for the parabola is the locus of a point on a plane that is equidistant from a fixed point called the focus and a fixed line called the directrix. Dandelin’s Proof connecting the two geometric definitions of a parabola. In this proof we will show that a when a plane intersects a cone at an angle that is parallel to the generator it will create a curve that is the locus of a point on a plane that is equidistant from a given line, the directrix, and a point, the focus. To begin we will first construct the parabola by cutting through the nappe of the cone with plane j at an angle that is parallel to the generator. Now, construct a sphere between the vertex of the cone, V, and the intersecting plane, such that the sphere is tangent to the cone at circle O, and plane j at point F the focus, simultaneously. The plane that contains circle O, plane k, will intersect with plane j, at a line called the directrix (see figure 22). Figure 22 Plane j V Plane k A B Circle O C Line d Directrix F P Parabola Copyright © 2000, 2007 Louis A. Talman Construct a line along the side of the cone from the vertex to point P, that intersects circle O, we will call this point of intersection point A (yellow line in figure 22). Now, construct a perpendicular line along plane j, to the directrix through point P (blue line in figure 22). We will also construct a perpendicular line to plane k through point P, we will call this point of intersection point C. In order to prove that the intersection of the cone and plane j is a parabola we need to show that for any point P on the parabola, PF PB. As point P moves on the parabola, points A will move around circle O, and point B will move along the directrix. We know that PF and PA are both - 13 - Samantha Klein 2009 tangents to the same sphere from and exterior point, thus PF and PA are congruent. Since PF = PA then we need to prove that PA PB, thus showing that PF PB. In order for us to show that PA PB , we need to prove that triangles PCA and PCB are congruent. Since we constructed line PC to be perpendicular to plane k, then we know that PCA PCB because they are both right angles. We also know that triangles PCA and PCB share a side PC , thus to prove the triangle are congruent we need to show that APC BPC . We know that line VT is the axis of the cone and is therefore perpendicular to plane k. We also know that line PC is perpendicular to plane, thus lines VT and PC are parallel, because they are perpendicular to the same plane (see figure 23). Figure 23 Plane j V S Plane k A B Circle O C Line d Directrix U T F P Parabola Copyright © 2000, 2007 Louis A. Talman Since VT and PC are parallel, TVA and APC are alternate interior angles and thus are congruent. Now construct a line along the side of the cone from the vertex to a point, we will call this point U. We can construct a parallel line, to line VU , on plane j that will intersect the directrix at a point S, and the axis of the cone at point T (see green lines in figure 23). Since VU is parallel to line ST we know STV TVU because they are alternate interior angles. We also know that TVU and TVA are congruent because lines VU and VP are both along the side of the cone. Since the sides of triangle BPC are parallel to those of triangle STV, it follows that BPC is congruent to STV . Since we know APC TVA and TVA TVU, then APC TVU . We also know that TVU STV and STV BPC, thus APC BPC . So by the angle side angle theorem we know that triangles BPC and APC are congruent. Therefore there corresponding parts are congruent so PA PB . Now since PA PB and PA PF , by the transitive property we know PF PB. Hence, a parabola is the intersection of a cone and a plane at an angle that is parallel to the generator, will create a curve that is the locus of a point on a plane that is equidistant from a given line, the directrix, and a point, the focus. Q.E.D. - 14 - Samantha Klein 2009 Derivation of standard algebraic form of the parabola. Using the locus definition for the parabola we can derive the standard form of the algebraic equation for the parabola. According to the definition a parabola is a locus of a point in a plane that is equidistant from a given point, the focus, and a given line, the directrix. To derive the standard form of the parabola we must start by constructing a line, the directrix and a given point, the focus. Then we need to construct the vertex of the parabola. The vertex of a parabola is defined as the lowest or highest point on the parabola. Because a parabola is a locus of a point that is equidistant from a point and line, we know that the vertex must be the midpoint between the focus and the directrix and must be on the axis of symmetry (see figure 24). Suppose the vertex of the parabola is at the origin (0,0), then we can place the Figure 24 directrix some distance, p, from the vertex. Let the equation for the directrix be y = -p. Since the vertex is the midpoint between the focus and directrix and we have placed the vertex at (0,0), then the focus must be located at (0,p). We know that the distance QuickTime™ and a decompressor from the focus to the directrix is now 2p. are needed(0, to see ) this picture. Now construct a line from any point (x, y) 2p on the curve to the focus, we will call this d1. Construct a line perpendicular to the directrix and passing through the point (x, y), this we will call d2.. Using the distance 2 2 formula the length of d1 x (y p) and the length of d2 y (p) y p. Now, since by definition a parabola is equidistant form the focus and directrix, we know 2 2 that d1 =d2 for any point (x, y) on the curve x (y p) y p. Now square both sides 2 to remove the radical x 2 (y p)2 y p 2 and we are left with x 2 y p 2 y p 2 . Multiply the squared terms and we have x 2 y 2 2yp p2 y 2 2yp p2 , thus x 2 4yp. This is the standard form of a parabola that has a vertex at the origin and opens up. We can also put this in a more common form x 2 of y . Where p is the distance to from the 4 p Figure 25 vertex to the focus. If the parabola where to open to the side then we can interchange x and y such the standard form would be y 2 4xp solve for y would leave us y 4xp . Now suppose the center was not at the origin but was at some point (h, k). Then the QuickTime™ and a directrix would be located at y = k –p, the axis of decompressor symmetry would be at x = h and the focus would be are needed(h,k+ to see ) this picture. located at (h, k +p) (see figure 25). Using the same derivation above would yield the standard form (h, k) x h 2 4 py k for a parabola having a vertical y =k-p - 15 - Samantha Klein 2009 axis of symmetry. For a parabola that has a horizontal axis of symmetry we would interchange the x and y and also the h and k, giving us the standard form y k 2 4 px h . Eccentricity Above we looked at conics as a locus of a point with a given parameter. But the parabola was the only locus definition given in terms of a focus and directrix. In fact all of the conics can be defined in a similar way using a focus and directrix. The focus- directrix property is what we think of today as a locus definition of all the conics. “Pappus’s enunciation of the theorem is to the effect that the locus of a point such that its distance from a fixed point is a given ratio to its distance from a fixed straight line is a conic” (Heath, 119). Today we refer to this ratio as the eccentricity of the conic. The eccentricity allows us to define all the conics in one way and to show the relationship between the conic sections. Each conic that we have discussed thus far can be defined in terms of its eccentricity. That is the ratio of the distance from a focus to a directrix. The parabola was defined as being equidistant from the focus to the directrix and therefore would have and eccentricity of 1. The circle is formed when the eccentricity is zero. Therefore the ellipse has an eccentricity that would be greater than zero but less than one (see figure 26). That is any eccentricity between that of a circle and a parabola. This is because the distance from the point to the focus is always less than the distance from the same point to the directrix on an ellipse. The hyperbola would have an eccentricity that is greater than one (see figure 27). This is because when the plane is less than the vertex angle the distance from a point to the focus is always greater than the distance from the same point to the directrix. Figure 27: Eccentricity of Hyperbola Figure 26: Eccentricity of ellipse Sides of cone Plane creating Sides of cone hyperbola Plane creating circle tangent to cone Distance to Distance to Distance to Focus (green Directrix Plane creating Directrix line) (red line) circle tangent to (red line) cone Plane creating Distance to ellipse Focus (green line) < so the ratio of < 1 > so the ratio of >1 This paper is just the start to a discussion on the conic sections. In this paper we looked at showing how the geometric definitions of the conics are related to on another, how eccentricity allows us to define all the conics in one definition and how to find the - 16 - Samantha Klein 2009 standard algebraic form of each conic section. The conics are a fascinating topic in mathematics that unifies the fundamentals of geometry and algebra. References 1. Belinsky, E. (n.d.). The Hyperbola. Retrieved from http://www.uz.ac.zw/science/maths/zimaths/hyperbol.htm 2. Bulmer-Thomas, I. (1939). Selections illustrating the history of Greek mathematics [Volume 2]. (Internet Archive), Retrieved from http://www.archive.org/details/selectionsillust02bulmuoft 3. David , H. (n.d.). Dandelin’s Spheres. Retrieved from http://math2.org/math/algebra/conics.htm 4. Heath, T. L. (1921). A history of Greek mathematics [Volume 2]. Oxford: The Clarendon Press. (google books version), Retrieved from http://books.google.com/books?id=7DDQAAAAMAAJ&dq=a%20history%20of%20 greek%20mathematics%20volume%202&pg=PP8#v=onepage&q&f=false 5. James,. (n.d.). 8.1 Conics. Retrieved from http://people.richland.edu/james/lecture/m116/conics/paradef.html 6. O'Connor, J.J, & Robertson, E.F. (2009, Janurary). Apollonius of Perga . Retrieved from http://www-history.mcs.st-andrews.ac.uk/Biographies/Apollonius.html 7. Picciotto, H. (n.d.). Geometry of the parabola (3d). Retrieved from http://www.MathEdPage.org/parabolas/geometry/conic.html 8. Schmarge, K. (1999). Conic sections in ancient Greece. Retrieved from http://www.math.rutgers.edu/~cherlin/History/Papers1999/schmarge.html 9. Talman, L.A. (2009, December 3). The Geometry of the conic sections. Retrieved from http://rowdy.mscd.edu/~talmanl/HTML/GeometryOfConicSections.html 10. Tuleja, ., & Hanc, J. (2002, Janurary). Java applet JDandelin. Retrieved from http://www.lostlecture.host.sk/JDandelinEn.htm 11. Conic sections. (n.d.). Retrieved from http://math2.org/math/algebra/conics.htm 12. Conics. (n.d.). Platonic Realms Interavtive Mathematics Encyclopedia, Retrieved from http://www.mathacademy.com/pr/prime/articles/conics/index.asp 13. Dandelin’s Spheres . (n.d.). Platonic Realms Interavtive Mathematics Encyclopedia, Retrieved from http://www.mathacademy.com/pr/prime/articles/dandelin/index.asp 14. SparkNotes Editors. (n.d.). SparkNote on Conic Sections. Retrieved April 1, 2010, from http://www.sparknotes.com/math/precalc/conicsections/ - 17 -