2003 ICTM Contest Division AAOrals Topic: Conics

Micah Fogel

Illinois Math and Science Academy

The Source

There is no official source this year.

• Source was to be Schaum’s Outline: Analytic Geometry

• Out of print!

Topics

• Locus definitions of conics

• Focus/directrix definitions of conics

• Rectangular equations

• Parametric equations

• Translation and rotation of axes

Topics (State level)

• Polar forms for conics

Additional Resources

• Mailing list: fogel@imsa.edu• Webstie: http://www.imsa.edu/~fogel• ICTM Website:

www.viebach.net/ictm/admin/c2003r.htm• Books

– Analytic geometry– Calculus– Last few orals sources

Study Ideas

• Study each aspect by itself (i.e., rectangular equations one day, proofs of reflection properties another day, etc.)

• Steal lots of basic exercises so that the algebra becomes automatic

• Try to find different explanations of the relationships between the various properties of conics

Cones and Conic Sections

Slicing a Cone

Intersect a double-cone with a plane. If the plane goes through the vertex of the cones, the intersection is either a point, a line, or two crossing lines.

Slicing a Cone

• These three cases are degenerate and are not really very interesting

• The interesting stuff happens when the plane does not go through the vertex

• Then there are three cases– Plane has less slant than cone– Plane has more slant than cone– Plane is parallel to slant of cone

Ellipse

• In the first case, there are two spheres, called Dandelin (Belgian, 1794–1847) spheres, which are tangent to both cone and plane

• The points where these spheres touch the plane will be called foci (sing.: focus)

Ellipse

• The intersection of the plane with the cone is called an ellipse

• Pick a point on the ellipse and draw the line through it and the vertex of the cone

• The distance between the two dashed circles on this line is constant

Ellipse

• But the distances to the foci are the same as the distances along this line, since they are tangents to the same spheres

• Thus, for any point on an ellipse, the sum of the distances to the foci is a constant

Ellipse

• Concentrate on one of the Dandelin spheres. Consider the plane through the circle where it is tangent to the cone

• Call the line where this plane intersects the cutting plane a directrix

Ellipse

• From a point P on the ellipse, travel parallel to the axis of the cone to the new plane, call this point Q. Travel along the cone until the intersection of the cone and sphere is reached. Call this point A. Choose D so that QD directrix

Q

P

A

F

D

Ellipse

• Note that in triangle PQD, we have PQ = PD cos , where is the angle between the cutting plane and the cone axis

P

DQ

Q

P

D

Ellipse

• Also note that angle QPA has the same measure as the angle between the axis and slant of the cone. So PQ = PA cos

• Also note PF = PA, since both are tangents to a sphere from P

Q

P

A

F

D

Ellipse

• Putting these relationships together gives PD cos = PF cos , or that

• Thus, since and are constants, the ratio of the distances from a point on the ellipse to a focus and to the corresponding directrix is a constant, called the eccentricity

€

PFPD

=cosαcosβ

Ellipse—Summary

• An ellipse is the intersection of a cone and a plane with less slant than the cone

• The ellipse has two foci, which are the intersection of the slant plane and the Dandelin spheres

• The sum of the distances from the foci to any point on the ellipse is a constant

Ellipse—Summary

• The slant plane meets the plane through the circle where a Dandelin sphere intersects the cone in a line called the directrix

• The distance from any point on the ellipse to a focus is a constant—the eccentricity— times the distance from that point to the directrix

• Since the plane has less slant than the cone, the eccentricity is less than one

Hyperbola

• If the slant of the plane is more than the slant of the cone, then the plane intersects both nappes of the cone

• This time, both Dandelin spheres are on the same side of the plane

• Thus, everything works the same, except that we have to subtract distances to foci to get a constant, and the eccentricity will be larger than one

Hyperbola—Summary

• A hyperbola is the intersection of a cone and a plane with more slant than the cone

• The hyperbola has two foci, which are the intersection of the slant plane and the Dandelin spheres

• The difference of the distances from the foci to any point on the hyperbola is a constant

Hyperbola—Summary

• The slant plane meets the plane through the circle where a Dandelin sphere intersects the cone in a line called the directrix

• The distance from any point on the hyperbola to a focus is a constant—the eccentricity— times the distance from that point to the directrix

• Since the plane has more slant than the cone, the eccentricity is greater than one

Parabola

• When the slant plane has the same slant as the cone itself, the intersection is a single curve called a parabola

• This time, there can only be one Dandelin sphere

• So a parabola has only one focus

Parabola

• We can repeat all the same arguments and calculations to find:– A parabola is the intersection of a cone with a plane

of the same slant– Because slants are equal, there is only one Dandelin

sphere, which meets the plane in the focus– Because slants are equal, eccentricity is one, so

parabola is the set of points equidistant from a line (the directrix) and a point (the focus)

Circle

• A special case of a plane intersecting a cone is if the plane is perpendicular to the axis of the cone, resulting in a circle

• If we repeat all the above constructions, there are two Dandelin spheres, which are both tangent to the cutting plane—circles have only one “focus,” the center

• The relevant planes are parallel, so circles have no directrices. Eccentricity is zero

Reflection Principle

• Given two points on the same side of a line, the shortest path from one to the other which touches the line meets the line at equal angles coming and going

• This can be seen by reflecting one point across the line, and knowing the shortest distance between the other point and the reflection is a straight line

Reflection Principle

P1

P2

P2’

Reflection Propety of Ellipses

• Apply this to the foci and a tangent line to an ellipse

• F1QF2 > F1PF2 since F1P’F2 = F1PF2, so P is the point at which incident angles are equal

F1 F2

P

Q

P’

Reflection Property of Parabolas

• If we move the focus F2 more and more to the right, the reflected ray becomes more and more horizontal

• If we let F2 “go to infinity” the ellipse turns into a parabola

• So a ray from the focus of a parabola will be reflected parallel to the axis of the parabola

Reflection Property of Hyperbolas

• If you keep moving F2 to the right, “past” infinity, it reappears on the left. The ellipse turns inside-out, and becomes a hyperbola

• So the corresponding reflection property for hyperbolas is that a ray from one focus will be reflected off the hyperbola directly away from the other focus(Of course, these aren’t rigorous proofs for parabola

and hyperbola!)

Reflection Property of Hyperbolas

Cartesian Equations for Conics

• We can use the distance formula and the focal distance properties of conics to find equations for their graphs in the Cartesian plane

• For instance, a circle with radius r and center (h, k) is simply

• Squaring, we obtain the standard equation

(x - h)2 + (y - k)2 = r2

€

(x − h)2 + (y − k)2 = r

Cartesian Equation for Ellipse

• If we put the foci at ( c, 0) and make the total distance of a point to the two foci 2a the distance formula gives us

• Move one radical to the other side, square, and simplify to obtain

€

(x − c)2 + y2 + (x + c)2 + y2 = 2a

€

−2cx = 2cx − 4a (x + c)2 + y2 + 4a2

Cartesian Equation for Ellipse

• Isolate the radical and square again:

a2(x2 + 2cx + c2 + y2) = a4 + 2a2cx + c2x2

• This simplifies to

(with b2 = a2 - c2)

• Eccentricity e = c/a

€

x2

a2 +y2

b2 =1

Cartesian Equation for Hyperbola

• The derivation for a hyperbola works exactly the same, except for the sign difference, yielding

(Foci are at (c, 0), difference between focal distances is 2a, and c2 = a2 + b2, eccentricity e = c/a)

€

x2

a2 −y2

b2 =1

Cartesian Equation for Parabola

• If we put the focus at (0, c) and the directrix as the line y = -c, we get the equation

• Squaring and simplifying, we obtain

x2 = 4cy• Naturally, we can translate any of these

equations to have center (h, k) by replaceing x with (x - h) and y with (y - k)

€

x2 + (y − c)2 = y + c

Info from the Equations

• We should be able to go from descriptions (“ellipse with its foci at (2, 5) and (2, 1) and eccentricity 0.4”) to Cartesian equations (here, center is (2, 3) so c = 2, c/a = 0.4 and a = 5, and therefore b2 = 52- 22 = 21, giving the equation (x - 2)2/21 +(y - 3)2/25 = 1)

• We should be able to go backward, finding center, foci, directrices, eccentricity, etc. from the Cartesian equation

General Quadratic Equation

• The general quadratic equation looks like Ax2 + Bxy + Cy2 + Dx + Ey + F = 0

• With no cross-product term, we could complete the square and reduce to one of our standard equations

• So we get rid of the cross-product term• If we replace x with x’cos - y’sin and

replace y with x’sin + y’cos we get….

General Quadratic Equations

A’x’2 + B’x’y’ + C’y’2 + D’x’ + E’y’ + F’ = 0

• We can obtain formulas for A’, B’, etc. The only important one is

B’ = Bcos(2 ) +(C - A)sin(2 )

• So if we set tan(2 ) = B/(A - C), B’ will equal zero

Rotation of Axes

• This process is called rotation of axes

• The new x’ and y’ axes are at angle to the original axes

• We can rotate the axes to eliminate the cross-product term, then complete the squares to reduce the equation to standard form

Example

Tell about 2x2 + 24xy + 9y2 +5x - 10y - 30 = 0

• First, if tan(2 ) = -24/7, cos( ) = 3/5 and sin( ) = 4/5

• So take x = 3x’/5 - 4y’/5, y = 4x’/5 + 3y’/5

• Substitute to obtain

18x’2 - 7y’2 - 5x’ - 10y’ - 30 = 0

Example

18(x’2 - 5/18 x’) - 7(y’2 - 10/7y’) - 30 = 0

18(x’2 - 5/18 x’ + 25/1296) - 7(y’2 - 10/7 y’ +25/49) - 30 - 25/72 + 25/7 = 0

18(x’ - 5/36)2 - 7(y’ - 5/7)2 = 16745/504

Discriminant

• The combination B2 - 4AC is left unchanged by any rotation of axes

• In particular, with the correct rotation, B = 0

• AC then tells whether you have an ellipse (both same sign), hyperbola (opposite sign) or parabola (one is zero)

• B2 - 4AC is called the discriminant. If negative, we have an ellipse. Positive gives a hyperbola. Zero gives a parabola.

Degenerate Conics

• If the discriminant is positive, the hyperbola may degenerate into a pair of crossing lines

• If the discriminant is negative, the ellipse might be a circle, a single point, or an empty graph

• If the discriminant is zero, the parabola may degenerate into a straight line, a pair of parallel lines, or an empty graph

Asymptotes

• Hyperbolas have asymptotes

• For the

asymptotes are found by setting the right-

hand side to 0• Then we get the lines y - k = (b/a)(x - h)

€

(x − h)2

a2 −(y − k)2

b2 = ±1

Parametric Equations

• Using the trig relations cos2 + sin2 = 1 and sec2 - tan2 = 1 we can transform any conic into a parametric equation

• Let x = h + a cos and y = k+ b sin . Then (x - h)/a = cos , (y - k)/b = sin , so we

have the ellipse

€

(x − h)2

a2 +(y − k)2

b2 =1

Parametric Equations

• Similarly, a general hyperbola might be given by x = h + tan , y = k + sec (for hyperbolas that open up and down) or by x = h + sec , y = k + tan (for hyperbolas that open to the sides)

• Parabolas can have y expressed as a function in x, or vice-versa, which are easy to turn into parametrics—let x (or y) be the parameter!

Polar Coordinates

• Using the directrix-focus properties of conics, it is easy to find equations for conics in polar coordinates.

• Recall x = r cos and y = r sin • Put a focus at the origin and make the

corresponding directrix the line x = p

• Recall that all conics have the distance from a point to the focus e (eccentricity) times the distance from the point to the directrix

Polar Coordinates

• So if (x, y) is on a conic we have that r equals e(x - p)

• Using x = r cos , we get r = e(r cos - p) or r(1 - e cos ) = -ep

• Thus the equation becomes

• We can rotate this to have different orientations by manipulating

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r =−ep

(1− ecosθ)