MA154: Algebra for 1st Year IT
Lecture 11/12: Conic Sections
Niall Madden
22 Mar 2007
CS457 — Lecture 11/12: Conic Sections
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A Conic Section is a curve formed by a plane
intersecting a cone.
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There are three types of conic sections:
y 2 = kx
x2 y2
+
=1
a2 b 2
x2 y2
−
=1
a2 b 2
CS457 — Lecture 11/12: Conic Sections
(Parabola);
(Ellipse);
(Hyperbola).
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Eccentricity
A Conic Section can also be described as
A set of all points (x, y ) such that the ratio of their
distance from a given point F = (p, 0) to their distance
from 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
1
This e is not “Euler’s Number” as in e iθ
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Eccentricity
Let PQ be the perpendicular from P to the line L, i.e.,
Q = (−p, y ).
p
Then |PF | = e |PQ| implies (x − p)2 + y 2 = e |x − (−p)| and
hence...
(x 2 − 2px + p 2 ) + y 2 = e 2 (x 2 + 2px + p 2 ) =⇒
x 2 (1 − e 2 ) − 2p(1 + e 2 )x + y 2 = −p 2 (1 − e 2 )
The cases
e = 1,
yield
a parabola,
respectively.
CS457 — Lecture 11/12: Conic Sections
e < 1,
an ellipse,
and
e>1
a hyperbola
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The Parabola
A Parabola is the set of all points P in the
plane that are equidistant from a fixed point
F , the focus of the parabola, and a fixed line
L, the parabola’s directrix, where L does not
contain F .
Standard equation: y 2 = 4px.
Focus: F = (p, 0). Directrix: x = −p.
The point of a parabola midway between its
focus and and its directrix is called the vertex
of the parabola.
The parabola y 2 = 4px has axis y = 0 and
vertex (0, 0).
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The Parabola
A parabola is symmetric about its axis, the line joining its focus to
its vertex.
That is because, if the parabola is defined as y 2 = 4px, this is the
same as (−y )2 = 4px
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The Parabola
Replacing x by −x;
interchanging x and y ;
replacing y by −y :
four different orientations in total.
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Examples of Parabolas
Example
Determine focus, directrix, axis and vertex of the parabola
x 2 = 12y .
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Examples of Parabolas
Example
Determine the graph of 4y 2 − 8x − 12y + 1 = 0.
Soln: (Complete the square:)
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Parabolae in the physical world
The trajectory of an object in motion under the influence of gravity
without air resistance is a parabola.
This was discovered by Galileo in the early 17th century, and
proven mathematically by Isaac Newton.
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Parabolae in the physical world
Approximations of parabolas are also found in the shape of cables
of suspension bridges.
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Parabolae in the physical world
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Parabolae in the physical world
Parabolic arches in
Antoni Gaudi’s Casa
Batllo in Barcelona.
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The Ellipse
Let 0 < e < 1, let F be a point and let L be a line not
containing F . The ellipse with eccentricity e, focus F and
directrix L is the set of all point P such that the distance |PF | is e
times 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 − b 2 .
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The Ellipse
One can also show that an ellipse is the set of points in the plane
whose distances from its two foci have a constant sum
q
q
2
2
(x + c) + y + (x − c)2 + y 2 = 2a.
Proof:
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The Ellipse
An ellipse is symmetric about two axes. Here a and b are the
lengths of the major and minor semiaxes, respectively.
If e = 0 then a = b (So a Circle is an ellipse with eccentricity 0).
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Ellipse Example
Example
Write down the equation of the ellipse with foci (±3, 0) and
vertices (±5, 0).
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Ellipse Example
Example
Determine the graph of x
3x 2 + 5y 2 − 12x + 30y + 42 = 0.
Soln: Collect terms and complete the square:
3(x 2 − 4x) + 5(y 2 + 6y ) = −42,
(x − 2)2 (y + 3)2
+
= 1.
5
3
This describes a translated ellipse with center
√ at (2, −3). Its
horizontal major semiaxis√has length a = 5 and its minor
semiaxis has length b = 3.
√
The distance form the center
p to each focus is c = 2 and the
eccentricity is e = c/a = 2/5.
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Recall .. the conics
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Hyperbola
Let e > 1, let F be a point and L a line not containing F . The
Hyperbola with eccentricity e, focus F and directrix L is the set
of all points P such that the distance |PF | is e times the distance
from P to the line L.
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Hyperbola
Standard equation:
x2
a2
−
y2
b2
= 1.
Foci: (±c, 0).
Directrices: x = ±a/e.
The center-to-focus distance is c =
√
a2 + b 2 .
A hyperbola is the set of points in the plane whose distances from
its two foci have a constant difference
q
q
(x + c)2 + y 2 − (x − c)2 + y 2 = ±2a
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Hyperbola
The points (±a, 0) are the vertices of the hyperbola. A hyperbola
is symmetric about two axes, the transverse axis joining the
vertices, and the conjugate axis.
A hyperbola has two branches. The lines y = ±bx/a passing
through the center (0, 0) are asymptotes of the two branches.
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Hyperbola Example
Example
Find the equation of the hyperbola with foci (±10, 0) and
asymptotes y = ±4x/3.
c = 10 and b/a = 4/3 with a2 + b 2 = c 2 give b = 8 and a = 6:
y2
x2
36 − 64 = 1.
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Hyperbola Example
Example
Determine the graph of the equation
9x 2 − 4y 2 − 36x + 8y = 4.
Collect terms and complete the square:
9(x − 2)2 − 4(y − 1)2 = 36,
(x − 2)2 (y − 1)2
−
= 1.
4
9
This describes a hyperbola with a horizontal transverse axis
√ and
centre (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 = ± 32 (x − 2).
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Quadratic Curves
With very few exceptions, the graph of a quadratic equation
Ax 2 + Bxy + Cy 2 + Dx + Ey + F = 0
is a conic section.
So far, the cross-product term Bxy has been absent due to
the fact that the axes of the conic sections ran parallel to the
coordinate axes.
To eliminate the xy -term we rotate the coordinate axes...
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Rotations...
The idea is to use a special linear transformation that rotate points
in R2 about the origin.
This can be done by:
changing from the coordinates (x, y ) to coordinates (x ? , y ? )
by...
the transformation
0
x
x
=A
y
y0
where A is an orthogonal matrix.
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Orthogonal Matrices
Definition A is orthogonal if its inverse is its transpose:
A−1 = AT .
a b
Example: If A =
then
c d
1
ad − bc
d −b
−c a
a c
b d
1 0
=
0 1
Example: For any angle α, If
cos α − sin α
A=
,
sin α cos α.
then A is Orthogonal.
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Orthogonal Matrices
Equations for a counterclockwise rotation about an angle α:
x = x 0 cos α − y 0 sin α,
y = x 0 sin α + y 0 cos α.
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Rotating a Quadratic Curve
Apply the rotation about α to the general quadratic equation
and obtain:
2
2
A 0x 0 + B 0x 0y 0 + C 0y 0 + D 0x 0 + E 0y 0 + F 0 = 0
where the coefficient of x 0 y 0 is
B 0 = B(cos2 α − sin2 α) + 2(C − A) sin α cos α
= B cos 2α + (C − A) sin 2α,
To find α with B 0 = 0, solve B cos 2α + (C − A) sin 2α = 0
for α, i.e.,
cot 2α =
CS457 — Lecture 11/12: Conic Sections
A−C
,
B
tan 2α =
B
.
A−C
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Rotating a Quadratic Curve
Example (Summer 2004, Q8(c))
Find an orthogonal transformation that reduces the conic
x 2 − 3xy + y 2 = 5 to standard form.
Solution: Here B = −3 and A − C = 0, so we use that
cot 2α = 0. This gives α = 3π/4
Now
3π
−1
3π
− y 0 sin
= √ (x 0 + y 0 )
4
4
2
3π
3π
1
y = x 0 sin
+ y 0 cos
= √ (x 0 − y 0 ).
4
4
2
x = x 0 cos
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Rotating a Quadratic Curve
Substitute back into the equation for the conic to get
3
1
1 0
(x + y 0 )2 + (x 0 + y 0 )(x 0 − y 0 ) + (x 0 − y 0 )2 = 5
2
2
2
3 1
3 1
0 2 1
0 0
0 2 1
(x ) ( + + ) + x y (0) + (y ) ( − + ) = 5
2 2 2
2 2 2
x 02 y 02
−
= 1.
2
10
√
√
So it√is a Hyperbola
a
=
2,
b
=
10 and centre-to-focus distance
√
c = 2 + 10 = 12.
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Classification
After removing the cross-product term Bxy , we consider an equation of
the form
Ax 2 + Cy 2 + Dx + Ey + F = 0.
(If the equation can be factored into two linear factors this describes one
or 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 at
all),
a hyperbola if A and C have opposite signs (or a pair of intersecting
lines),
a straight line if A = C = 0 and at least one of D and E is not 0.
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The Discriminant
The discriminant of the quadratic equation is the number
B 2 − 4AC .
The discriminant does not change under rotation:
2
B 0 − 4A 0 C 0 = B 2 − 4AC
The curve is a parabola if B 2 − 4AC = 0.
The curve is an ellipse if B 2 − 4AC < 0.
The curve is a hyperbola if B 2 − 4AC > 0.
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