MA154: Algebra for 1st Year IT

Lecture 11/12: Conic Sections

Niall Madden

22 Mar 2007

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A Conic Section is a curve formed by a planeintersecting a cone.

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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).

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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 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

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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.

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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).

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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

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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 parabolax2 = 12y .

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Examples of Parabolas

Example

Determine the graph of 4y2 − 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 gravitywithout air resistance is a parabola.This was discovered by Galileo in the early 17th century, andproven mathematically by Isaac Newton.

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Parabolae in the physical world

Approximations of parabolas are also found in the shape of cablesof suspension bridges.

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Parabolae in the physical world

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Parabolae in the physical world

Parabolic arches inAntoni Gaudi’s CasaBatllo in Barcelona.

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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.

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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:

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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).

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Ellipse Example

Example

Write down the equation of the ellipse with foci (±3, 0) andvertices (±5, 0).

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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.

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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 . 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.

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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 + 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

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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.

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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.

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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).

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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...

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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 .

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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.

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Orthogonal Matrices

Equations for a counterclockwise rotation about an angle α:

x = x ′ cosα− y ′ sinα,y = x ′ sinα+ y ′ cosα.

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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.

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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 ′).

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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.

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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.

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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.

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