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MA154: Algebra for 1st Year IT
Lecture 11/12: Conic Sections
Niall Madden
22 Mar 2007
CS457 — Lecture 11/12: Conic Sections 1/34
A Conic Section is a curve formed by a planeintersecting a cone.
CS457 — Lecture 11/12: Conic Sections 2/34
There are three types of conic sections:
y 2 = kx (Parabola);
x2
a2+
y 2
b2= 1 (Ellipse);
x2
a2−
y 2
b2= 1 (Hyperbola).
CS457 — Lecture 11/12: Conic Sections 3/34
Eccentricity
A Conic Section can also be described as
A set of all points (x , y) such that the ratio of theirdistance from a given point F = (p, 0) to their distancefrom the line L with equation x = −p is a fixed ration e.
This number e > 0 is called the Eccentricity of a the Conic.1
1This e is not “Euler’s Number” as in e iθ
CS457 — Lecture 11/12: Conic Sections 4/34
Eccentricity
Let PQ be the perpendicular from P to the line L, i.e.,Q = (−p, y).Then |PF | = e |PQ | implies
√(x − p)2 + y2 = e |x − (−p)| and
hence...
(x2 − 2px + p2) + y2 = e2(x2 + 2px + p2) =⇒x2(1 − e2) − 2p(1 + e2)x + y2 = −p2(1 − e2)
The cases e = 1, e < 1, and e > 1yield a parabola, an ellipse, a hyperbolarespectively.
CS457 — Lecture 11/12: Conic Sections 5/34
The Parabola
A Parabola is the set of all points P in theplane that are equidistant from a fixed pointF , the focus of the parabola, and a fixed lineL, the parabola’s directrix , where L does notcontain F .Standard equation: y2 = 4px .Focus: F = (p, 0). Directrix: x = −p.The point of a parabola midway between itsfocus and and its directrix is called the vertexof the parabola.
The parabola y2 = 4px has axis y = 0 and
vertex (0, 0).
CS457 — Lecture 11/12: Conic Sections 6/34
The Parabola
A parabola is symmetric about its axis, the line joining its focus toits vertex.That is because, if the parabola is defined as y2 = 4px , this is thesame as (−y)2 = 4px
CS457 — Lecture 11/12: Conic Sections 7/34
The Parabola
Replacing x by −x ;
interchanging x and y ;
replacing y by −y :
four different orientations in total.
CS457 — Lecture 11/12: Conic Sections 8/34
Examples of Parabolas
Example
Determine focus, directrix, axis and vertex of the parabolax2 = 12y .
CS457 — Lecture 11/12: Conic Sections 9/34
Examples of Parabolas
Example
Determine the graph of 4y2 − 8x − 12y + 1 = 0.
Soln: (Complete the square:)
CS457 — Lecture 11/12: Conic Sections 10/34
Parabolae in the physical world
The trajectory of an object in motion under the influence of gravitywithout air resistance is a parabola.This was discovered by Galileo in the early 17th century, andproven mathematically by Isaac Newton.
CS457 — Lecture 11/12: Conic Sections 11/34
Parabolae in the physical world
Approximations of parabolas are also found in the shape of cablesof suspension bridges.
CS457 — Lecture 11/12: Conic Sections 12/34
Parabolae in the physical world
CS457 — Lecture 11/12: Conic Sections 13/34
Parabolae in the physical world
Parabolic arches inAntoni Gaudi’s CasaBatllo in Barcelona.
CS457 — Lecture 11/12: Conic Sections 14/34
The Ellipse
Let 0 < e < 1, let F be a point and let L be a line notcontaining F . The ellipse with eccentricity e, focus F anddirectrix L is the set of all point P such that the distance |PF | is etimes the distance from P to the line L.
Standard equation: x2
a2 + y2
b2 = 1.
Foci: (±c , 0).
Directrices: x = ±a/e.
The points (±a, 0) are the vertices of the ellipse.The Center-To-Focus Distance is c =
√a2 − b2.
CS457 — Lecture 11/12: Conic Sections 15/34
The Ellipse
One can also show that an ellipse is the set of points in the planewhose distances from its two foci have a constant sum√
(x + c)2 + y2 +
√(x − c)2 + y2 = 2a.
Proof:
CS457 — Lecture 11/12: Conic Sections 16/34
The Ellipse
An ellipse is symmetric about two axes. Here a and b are thelengths of the major and minor semiaxes, respectively.If e = 0 then a = b (So a Circle is an ellipse with eccentricity 0).
CS457 — Lecture 11/12: Conic Sections 17/34
Ellipse Example
Example
Write down the equation of the ellipse with foci (±3, 0) andvertices (±5, 0).
CS457 — Lecture 11/12: Conic Sections 18/34
Ellipse Example
Example
Determine the graph of x
3x2 + 5y2 − 12x + 30y + 42 = 0.
Soln: Collect terms and complete the square:
3(x2 − 4x) + 5(y2 + 6y) = −42,
(x − 2)2
5+
(y + 3)2
3= 1.
This describes a translated ellipse with center at (2, −3). Itshorizontal major semiaxis has length a =
√5 and its minor
semiaxis has length b =√
3.The distance form the center to each focus is c =
√2 and the
eccentricity is e = c/a =√
2/5.
CS457 — Lecture 11/12: Conic Sections 19/34
Recall .. the conics
CS457 — Lecture 11/12: Conic Sections 20/34
Hyperbola
Let e > 1, let F be a point and L a line not containing F . TheHyperbola with eccentricity e, focus F and directrix L is the setof all points P such that the distance |PF | is e times the distancefrom P to the line L.
CS457 — Lecture 11/12: Conic Sections 21/34
Hyperbola
Standard equation: x2
a2 − y2
b2 = 1.
Foci: (±c , 0).
Directrices: x = ±a/e.
The center-to-focus distance is c =√
a2 + b2.
A hyperbola is the set of points in the plane whose distances fromits two foci have a constant difference√
(x + c)2 + y2 −
√(x − c)2 + y2 = ±2a
CS457 — Lecture 11/12: Conic Sections 22/34
Hyperbola
The points (±a, 0) are the vertices of the hyperbola. A hyperbolais symmetric about two axes, the transverse axis joining thevertices, and the conjugate axis.A hyperbola has two branches. The lines y = ±bx/a passingthrough the center (0, 0) are asymptotes of the two branches.
CS457 — Lecture 11/12: Conic Sections 23/34
Hyperbola Example
Example
Find the equation of the hyperbola with foci (±10, 0) andasymptotes y = ±4x/3.
c = 10 and b/a = 4/3 with a2 + b2 = c2 give b = 8 and a = 6:x2
36 − y2
64 = 1.
CS457 — Lecture 11/12: Conic Sections 24/34
Hyperbola Example
Example
Determine the graph of the equation
9x2 − 4y2 − 36x + 8y = 4.
Collect terms and complete the square:
9(x − 2)2 − 4(y − 1)2 = 36,
(x − 2)2
4−
(y − 1)2
9= 1.
This describes a hyperbola with a horizontal transverse axis andcentre (2, 1). From a = 2 and b = 3, it follows that c =
√13.
The vertices are (0, 1) and (4, 1) and the foci are (2±√
13, 1).The asymptotes are y − 1 = ±3
2(x − 2).
CS457 — Lecture 11/12: Conic Sections 25/34
Quadratic Curves
With very few exceptions, the graph of a quadratic equation
Ax2 + Bxy + Cy2 + Dx + Ey + F = 0
is a conic section.
So far, the cross-product term Bxy has been absent due tothe fact that the axes of the conic sections ran parallel to thecoordinate axes.
To eliminate the xy -term we rotate the coordinate axes...
CS457 — Lecture 11/12: Conic Sections 26/34
Rotations...
The idea is to use a special linear transformation that rotate pointsin R2 about the origin.This can be done by:
changing from the coordinates (x , y) to coordinates (x?, y?)
by...
the transformation (xy
)= A
(x ′
y ′
)where A is an orthogonal matrix .
CS457 — Lecture 11/12: Conic Sections 27/34
Orthogonal Matrices
Definition A is orthogonal if its inverse is its transpose:A−1 = AT .
Example: If A =
(a bc d
)then
1
ad − bc
(d −b
−c a
)(a cb d
)=
(1 00 1
)Example: For any angle α, If
A =
(cosα − sinαsinα cosα.
),
then A is Orthogonal.
CS457 — Lecture 11/12: Conic Sections 28/34
Orthogonal Matrices
Equations for a counterclockwise rotation about an angle α:
x = x ′ cosα− y ′ sinα,y = x ′ sinα+ y ′ cosα.
CS457 — Lecture 11/12: Conic Sections 29/34
Rotating a Quadratic Curve
Apply the rotation about α to the general quadratic equationand obtain:
A ′x ′2 + B ′x ′y ′ + C ′y ′2 + D ′x ′ + E ′y ′ + F ′ = 0
where the coefficient of x ′y ′ is
B ′ = B(cos2 α− sin2 α) + 2(C − A) sinα cosα
= B cos 2α+ (C − A) sin 2α,
To find α with B ′ = 0, solve B cos 2α+ (C − A) sin 2α = 0for α, i.e.,
cot 2α =A − C
B, tan 2α =
B
A − C.
CS457 — Lecture 11/12: Conic Sections 30/34
Rotating a Quadratic Curve
Example (Summer 2004, Q8(c))
Find an orthogonal transformation that reduces the conicx2 − 3xy + y2 = 5 to standard form.
Solution: Here B = −3 and A − C = 0, so we use thatcot 2α = 0. This gives α = 3π/4Now
x = x ′ cos3π
4− y ′ sin
3π
4=
−1√2(x ′ + y ′)
y = x ′ sin3π
4+ y ′ cos
3π
4=
1√2(x ′ − y ′).
CS457 — Lecture 11/12: Conic Sections 31/34
Rotating a Quadratic Curve
Substitute back into the equation for the conic to get
1
2(x ′ + y ′)2 +
3
2(x ′ + y ′)(x ′ − y ′) +
1
2(x ′ − y ′)2 = 5
(x ′)2(1
2+
3
2+
1
2) + x ′y ′(0) + (y ′)2(
1
2−
3
2+
1
2) = 5
x ′2
2−
y ′2
10= 1.
So it is a Hyperbola a =√
2, b =√
10 and centre-to-focus distancec =√
2 + 10 =√
12.
CS457 — Lecture 11/12: Conic Sections 32/34
Classification
After removing the cross-product term Bxy , we consider an equation ofthe form
Ax2 + Cy2 + Dx + Ey + F = 0.
(If the equation can be factored into two linear factors this describes oneor two straight lines. Otherwise) this is
a circle if A = C 6= 0 (or a single point or no graph at all),
a parabola if one of A or C is 0,
an ellipse if A and C have the same sign (or a point or no graph atall),
a hyperbola if A and C have opposite signs (or a pair of intersectinglines),
a straight line if A = C = 0 and at least one of D and E is not 0.
CS457 — Lecture 11/12: Conic Sections 33/34
The Discriminant
The discriminant of the quadratic equation is the number
B2 − 4AC .
The discriminant does not change under rotation:
B ′2 − 4A ′C ′ = B2 − 4AC
The curve is a parabola if B2 − 4AC = 0.
The curve is an ellipse if B2 − 4AC < 0.
The curve is a hyperbola if B2 − 4AC > 0.
CS457 — Lecture 11/12: Conic Sections 34/34