aarjaneiro/planetmath huggingface dataset:
intersection of quadratic surface and plane

intersection of quadratic surface and plane

The intersection of a sphere with a plane (

http://planetmath.org/IntersectionOfSphereAndPlane

) is a circle, similarly the intersection of any

surface of revolution

formed by the revolution of an

ellipse

or a

hyperbola

about its axis with a plane

perpendicular

to the axis of revolution is a circle of latitude.

We can get as intersection curves of other

quadratic surfaces

and a plane also other

quadratic curves

(conics).  If for example the

ellipsoid

x

2

a

2

+

y

2

b

2

+

z

2

c

2

=

1

(1)

is cut with the plane

z

=

0

(i.e. the

x

⁢

y

-plane), we substitute

z

=

0

to the equation of the ellipsoid, and thus the intersection curve satisfies the equation

x

2

a

2

+

y

2

b

2

=

1

,

which an ellipse.  Actually, all plane intersections of the ellipsoid are ellipses, which may be in special cases circles.

As another exaple of quadratic surface we take the

hyperbolic paraboloid

x

2

a

2

-

y

2

b

2

=

2

⁢

z

.

(2)

Cutting it e.g. with the plane

y

=

b

,  which is

parallel

to the

z

⁢

x

-plane, the substitution yields the equation

2

⁢

z

=

x

2

a

2

-

1

meaning that the intersection curve in the plane

y

=

b

has the

projection

(

http://planetmath.org/ProjectionOfPoint

)

parabola

in the

z

⁢

x

-plane with such an equation, and accordingly is such a parabola.

If we cut the

surface

(2) with the plane

z

=

1

2

, the result is the hyperbola
having the projection

x

2

a

2

-

y

2

b

2

=

1

in the

x

⁢

y

-plane.  But cutting with

z

=

0

gives

x

2

a

2

-

y

2

b

2

=

0

, i.e. the pair of
lines

y

=

±

b

a

⁢

x

which is a degenerate conic.

Let us then consider the general equation

A

⁢

x

2

+

B

⁢

y

2

+

C

⁢

z

2

+

2

⁢

A

′

⁢

y

⁢

z

+

2

⁢

B

′

⁢

z

⁢

x

+

2

⁢

C

′

⁢

x

⁢

y

+

2

⁢

A

′′

⁢

x

+

2

⁢

B

′′

⁢

y

+

2

⁢

C

′′

⁢

z

+

D

=

0

(3)

of quadratic surface and an arbitrary plane

a

⁢

x

+

b

⁢

y

+

c

⁢

z

+

d

=

0

(4)

where at least one of the

coefficients

a

,

b

,

c

is distinct from zero.  Their intersection equation is obtained, supposing that e.g.

c

≠

0

, by substituting the solved form

z

=

-

a

⁢

x

+

b

⁢

y

+

d

c

of (4) to the equation (3).  We then apparently have the equation of the form

α

⁢

x

2

+

β

⁢

y

2

+

2

⁢

γ

⁢

x

⁢

y

+

2

⁢

δ

⁢

x

+

2

⁢

ε

⁢

y

+

ζ

=

0

,

which a quadratic curve (

http://planetmath.org/QuadraticCurves

) or some of the degenerated cases of them.

Title

intersection of quadratic surface and plane

Canonical name

IntersectionOfQuadraticSurfaceAndPlane

Date of creation

2013-03-22 18:31:38

Last modified on

2013-03-22 18:31:38

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

7

Author

pahio (2872)

Entry type

Result

Classification

msc 51N20

Related topic

QuadraticSurfaces

Related topic

QuadraticCurves

Related topic

Conic

Related topic

EquationOfPlane

Related topic

ConicSection

Generated on Fri Feb 9 22:13:09 2018 by

LaTeXML
isogonal trajectory

isogonal trajectory

Let a one-parametric family of

plane curves

γ

have the

differential equation

F

⁢

(

x

,

y

,

d

⁢

y

d

⁢

x

)

=

0

.

(1)

We want to determine the

isogonal trajectories

of this family, i.e. the curves

ι

intersecting all members of the family under a given angle, which is denoted by

ω

.
For this purpose, we denote the

slope angle

of any curve

γ

at such an intersection point by

α

and the slope angle of

ι

at the same point by

β

.  Then

β

-

α

=

ω

(

or alternatively

-

ω

)

,

and accordingly

d

⁢

y

d

⁢

x

=

tan

⁡

α

=

tan

⁡

β

-

tan

⁡

ω

1

+

tan

⁡

β

⁢

tan

⁡

ω

=

y

′

-

tan

⁡

ω

1

+

y

′

⁢

tan

⁡

ω

,

where

y

′

means the slope of

ι

.  Thus the equation

F

⁢

(

x

,

y

,

y

′

-

tan

⁡

ω

1

+

y

′

⁢

tan

⁡

ω

)

=

0

(2)

is satisfied by the derivative

y

′

of the

ordinate

of

ι

.  In other , (2) is the differential equation of all isogonal trajectories of the given

family of curves

.

Note.

In the special case

ω

=

π

2

,  it’s a question of

orthogonal trajectories

.

Title

isogonal trajectory

Canonical name

IsogonalTrajectory

Date of creation

2013-03-22 18:59:20

Last modified on

2013-03-22 18:59:20

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

7

Author

pahio (2872)

Entry type

Derivation

Classification

msc 51N20

Classification

msc 34A26

Classification

msc 34A09

Related topic

AngleBetweenTwoCurves

Related topic

OrthogonalCurves

Related topic

ExampleOfIsogonalTrajectory

Related topic

AngleBetweenTwoLines

Defines

isogonal trajectory

Generated on Fri Feb 9 22:13:11 2018 by

LaTeXML
octant

octant

The

coordinate planes

(

y

⁢

z

-plane,

z

⁢

x

-plane,

x

⁢

y

-plane) the space

ℝ

3

into eight trihedral angles, which are called the

octants

of the space.  The magnitude of each octant as a

solid angle

is

π

2

.

Any

coordinate

of a point moving in a certain octant

preserves

its sign (

http://planetmath.org/SignumFunction

).  One calls the octant, where all three coordinates are nonnegative,

the

first octant

.  The other octants have no agreed numbering; see still

http://de.wikipedia.org/wiki/Oktant_(Geometrie)

Wikipedia.

Title

octant

Canonical

name

Octant

Date of creation

2013-03-22 18:51:53

Last modified on

2013-03-22 18:51:53

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

5

Author

pahio (2872)

Entry type

Definition

Classification

msc 51N20

Related topic

Quadrant

Defines

first octant

Generated on Fri Feb 9 22:13:13 2018 by

LaTeXML
parallelism of two planes

parallelism of two planes

Two planes

π

and

ϱ

in the 3-dimensional

Euclidean space

are

parallel

iff they either have no common points or coincide, i.e. iff

π

∩

ϱ

=

∅

or

π

∩

ϱ

=

π

.

(1)

An

equivalent

(

http://planetmath.org/Equivalent3

) condition of the parallelism is that the normal vectors of

π

and

ϱ

are parallel.

The parallelism of planes is an

equivalence relation

in any set of planes of the space.

If the planes have the equations

A

1

⁢

x

+

B

1

⁢

y

+

C

1

⁢

z

+

D

1

=

0

and

A

2

⁢

x

+

B

2

⁢

y

+

C

2

⁢

z

+

D

2

=

0

,

(2)

the parallelism means the proportionality (

http://planetmath.org/

Variation

) of the

coefficients

of the

variables

:  there exists a

k

such that

A

1

=

k

⁢

A

2

,

B

1

=

k

⁢

B

2

,

C

1

=

k

⁢

C

2

.

(3)

In this case, if also

D

1

=

k

⁢

D

2

,  then the planes coincide.

Using vectors, the condition (3) may be written

(

A

1

B

1

C

1

)

=

k

⁢

(

A

2

B

2

C

2

)

(4)

which equation utters the parallelism (

http://planetmath.org/MutualPositionsOfVectors

) of the normal vectors.

Remark.

The shortest

distance

of the

parallel planes

A

⁢

x

+

B

⁢

y

+

C

⁢

z

+

D

=

0

and

A

⁢

x

+

B

⁢

y

+

C

⁢

z

+

E

=

0

is obtained from the

d

=

|

D

-

E

|

A

2

+

B

2

+

C

2

,

(5)

as is easily shown by using Lagrange multipliers (

http://planetmath.org/LagrangeMultiplierMethod

) (see

http://planetmath.org/node/11604

this entry).

Title

parallelism of two planes

Canonical name

ParallelismOfTwoPlanes

Date of creation

2013-03-22 18:48:10

Last modified on

2013-03-22 18:48:10

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

15

Author

pahio (2872)

Entry type

Topic

Classification

msc 51N20

Classification

msc 51M04

Classification

msc 51A05

Synonym

parallelism of planes

Synonym

parallel planes

Related topic

PlaneNormal

Related topic

ParallelAndPerpendicularPlanes

Related topic

ParallelityOfLineAndPlane

Related topic

ExampleOfUsingLagrangeMultipliers

Related topic

NormalOfPlane

Defines

parallel

Defines

parallelism

Generated on Fri Feb 9 22:13:28 2018 by

LaTeXML
pencil of lines

pencil of lines

Let

A

i

⁢

x

+

B

i

⁢

y

+

C

i

=

0

(1)

be equations of some lines.  Use the short notations

A

i

⁢

x

+

B

i

⁢

y

+

C

i

:=

L

i

.

If the lines

L

1

=

0

and

L

2

=

0

have an

intersection

point

P

, then, by the parent entry (

http://planetmath.org/LineThroughAnIntersectionPoint

), the equation

k

1

⁢

L

1

+

k

2

⁢

L

2

=

0

(2)

with various real values of

k

1

and

k

2

can any line

passing through

the point

P

; this set of lines is called a

pencil of lines

.

Theorem

.

A necessary and

sufficient

condition in to three lines

L

1

=

0

,

L

2

=

0

,

L

3

=

0

pass through a same point, is that the

determinant

formed by the

coefficients

of their equations (1)

vanishes

:

|

A

1

B

1

C

1

A

2

B

2

C

2

A

3

B

3

C

3

|

=

|

A

1

A

2

A

3

B

1

B

2

B

3

C

1

C

2

C

3

|

=

0

.

Proof.

If the line

L

3

=

0

belongs to the fan of lines determined by the lines

L

1

=

0

and

L

2

=

0

,  i.e. all the three lines have a common point, there must be the

identity

L

3

≡

L

1

+

L

2

,

i.e. there exist three

real numbers

k

1

,

k

2

,

k

3

, which are not all zeroes, such that the equation

k

1

⁢

L

1

+

k

2

⁢

L

2

+

k

3

⁢

L

3

≡

0

(3)

is satisfied identically by all real values of

x

and

y

.
This means that the group of

homogeneous

linear equations

{

k

1

⁢

A

1

+

k

2

⁢

A

2

+

k

3

⁢

A

3

=

0

,

k

1

⁢

B

1

+

k

2

⁢

B

2

+

k

3

⁢

B

3

=

0

,

k

1

⁢

C

1

+

k

2

⁢

C

2

+

k

3

⁢

C

3

=

0

has nontrivial solutions

k

1

,

k

2

,

k

3

.
By

linear algebra

, it follows that the determinant of this group of equations has to vanish.

Suppose

conversely

that the determinant vanishes.  This implies that the above group of equations has a nontrivial solution

k

1

,

k

2

,

k

3

.  Thus we can write the identic equation (3).  Let e.g.

k

1

≠

0

.  Solving (3) for

L

1

yields

L

1

≡

-

k

2

⁢

L

2

+

k

3

⁢

L

3

k

1

,

which shows that the line

L

1

=

0

belongs to the fan determined by the lines

L

2

=

0

and

L

3

=

0

; so the lines pass through a common point.

References

1

Lauri Pimiä

:

Analyyttinen geometria

.  Werner Söderström Osakeyhtiö, Porvoo and Helsinki (1958).

Title

pencil of lines

Canonical name

PencilOfLines

Date of creation

2013-03-22 18:09:03

Last modified on

2013-03-22 18:09:03

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

8

Author

pahio (2872)

Entry type

Theorem

Classification

msc 51N20

Related topic

LineInThePlane

Related topic

Determinant2

Related topic

HomogeneousLinearProblem

Related topic

Pencil2

Generated on Fri Feb 9 22:13:34 2018 by

LaTeXML
proof of area of a quadrilateral

proof of area of a quadrilateral

One should think of the following diagram as representing a general

quadrilateral

.

Generated on Fri Feb 9 22:13:44 2018 by

LaTeXML
proof of Steiner’s theorem

proof of Steiner’s theorem

Using

α

,

β

,

γ

,

δ

to denote angles as in the diagram
at left, the sines law yields

A

⁢

B

sin

⁡

(

γ

)

=

A

⁢

C

sin

⁡

(

β

)

(1)

N

⁢

B

sin

⁡

(

α

+

δ

)

=

N

⁢

A

sin

⁡

(

β

)

(2)

M

⁢

C

sin

⁡

(

α

+

δ

)

=

M

⁢

A

sin

⁡

(

γ

)

(3)

M

⁢

B

sin

⁡

(

α

)

=

M

⁢

A

sin

⁡

(

β

)

(4)

N

⁢

C

sin

⁡

(

α

)

=

N

⁢

A

sin

⁡

(

γ

)

(5)

Dividing (2) and (3), and (4) by (5):

M

⁢

A

N

⁢

A

⁢

N

⁢

B

M

⁢

C

=

sin

⁡

(

γ

)

sin

⁡

(

β

)

=

N

⁢

A

M

⁢

A

⁢

M

⁢

B

N

⁢

C

and therefore

N

⁢

B

⋅

M

⁢

B

M

⁢

C

⋅

N

⁢

C

=

sin

2

⁡

(

γ

)

sin

2

⁡

(

β

)

=

A

⁢

B

2

A

⁢

C

2

by (1).

Title

proof of Steiner’s theorem

Canonical name

ProofOfSteinersTheorem

Date of creation

2013-03-22 13:48:06

Last modified on

2013-03-22 13:48:06

Owner

mathcam (2727)

Last modified by

mathcam (2727)

Numerical id

5

Author

mathcam (2727)

Entry type

Proof

Classification

msc 51N20

Generated on Fri Feb 9 22:13:46 2018 by

LaTeXML
proof of Van Aubel theorem

proof of Van Aubel theorem

We want to prove

C

⁢

P

P

⁢

F

=

C

⁢

D

D

⁢

B

+

C

⁢

E

E

⁢

A

On the picture, let us call

ϕ

to the angle

∠

⁢

A

⁢

B

⁢

E

and

ψ

to the angle

∠

⁢

E

⁢

B

⁢

C

.

A

generalization

of

bisector

’s

theorem

states

C

⁢

E

E

⁢

A

=

C

⁢

B

⁢

sin

⁡

ψ

A

⁢

B

⁢

sin

⁡

ϕ

on

⁢

△

⁢

A

⁢

B

⁢

C

and

C

⁢

P

P

⁢

F

=

C

⁢

B

⁢

sin

⁡

ψ

F

⁢

B

⁢

sin

⁡

ϕ

on

⁢

△

⁢

F

⁢

B

⁢

C

.

From the two equalities we can get

C

⁢

E

⋅

A

⁢

B

E

⁢

A

=

C

⁢

P

⋅

F

⁢

B

P

⁢

F

and thus

C

⁢

P

P

⁢

F

=

C

⁢

E

⋅

A

⁢

B

E

⁢

A

⋅

F

⁢

B

.

Since

A

⁢

B

=

A

⁢

F

+

F

⁢

B

, substituting leads to

C

⁢

E

⋅

A

⁢

B

E

⁢

A

⋅

F

⁢

B

=

C

⁢

E

⁢

(

A

⁢

F

+

F

⁢

B

)

E

⁢

A

⋅

F

⁢

B

=

C

⁢

E

⋅

A

⁢

F

E

⁢

A

⋅

F

⁢

B

+

C

⁢

E

⋅

F

⁢

B

E

⁢

A

⋅

F

⁢

B

=

C

⁢

E

⋅

A

⁢

F

E

⁢

A

⋅

F

⁢

B

+

C

⁢

E

E

⁢

A

But Ceva’s theorem states

C

⁢

E

E

⁢

A

⋅

A

⁢

F

F

⁢

B

⋅

B

⁢

D

D

⁢

C

=

1

and so

C

⁢

E

⋅

A

⁢

F

E

⁢

A

⋅

F

⁢

B

=

C

⁢

D

D

⁢

B

Subsituting the last equality gives the desired result.

Title

proof of Van Aubel theorem

Canonical name

ProofOfVanAubelTheorem

Date of creation

2013-03-22 14:03:28

Last modified on

2013-03-22 14:03:28

Owner

drini (3)

Last modified by

drini (3)

Numerical id

6

Author

drini (3)

Entry type

Proof

Classification

msc 51N20

Synonym

Van Aubel’s theorem

Related topic

ProofOfVanAubelsTheorem

Related topic

CevasTheorem

Related topic

VanAubelTheorem

Related topic

TrigonometricVersionOfCevasTheorem

Generated on Fri Feb 9 22:13:48 2018 by

LaTeXML
proof of Van Aubel’s theorem

proof of Van Aubel’s theorem

As in the figure, let us denote by

u

,

v

,

w

,

x

,

y

,

z

the areas of the
six

component

triangles

.
Given any two triangles of the same height, their areas are in the
same proportion as their bases (Euclid VI.1). Therefore

y

+

z

x

=

u

+

v

w

w

+

x

v

=

y

+

z

u

u

+

v

z

=

w

+

x

y

and the

conclusion

we want is

y

+

z

+

u

v

+

w

+

x

+

z

+

u

+

v

w

+

x

+

y

=

y

+

z

x

.

Clearing the denominators, the hypotheses are

w

⁢

(

y

+

z

)

=

x

⁢

(

u

+

v

)

(1)

y

⁢

(

u

+

v

)

=

z

⁢

(

w

+

x

)

(2)

u

⁢

(

w

+

x

)

=

v

⁢

(

y

+

z

)

(3)

which imply

v

⁢

x

⁢

z

=

u

⁢

w

⁢

y

(4)

and the conclusion says that

x

(

w

⁢

y

¯

+

w

⁢

z

¯

+

u

w

+

x

⁢

y

¯

+

x

⁢

z

¯

+

u

x

+

y

2

¯

+

y

⁢

z

¯

+

u

y

+

v

z

+

u

v

+

v

2

+

w

z

+

u

w

+

v

w

+

x

z

+

u

x

+

v

x

)

equals

(

y

+

z

)

⁢

(

v

⁢

w

+

v

⁢

x

+

v

⁢

y

+

w

2

+

w

⁢

x

+

w

⁢

y

+

w

⁢

x

¯

+

x

2

¯

+

x

⁢

y

¯

)

or equivalently (after cancelling the underlined terms)

x

⁢

(

u

⁢

w

+

x

⁢

z

+

u

⁢

x

+

u

⁢

y

+

v

⁢

z

+

u

⁢

v

+

v

2

+

w

⁢

z

+

u

⁢

w

+

v

⁢

w

+

u

⁢

x

+

v

⁢

x

)

equals

(

y

+

z

)

⁢

(

v

⁢

w

+

v

⁢

x

+

v

⁢

y

+

w

2

+

w

⁢

x

+

w

⁢

y

)

=

(

y

+

z

)

⁢

(

v

+

w

)

⁢

(

w

+

x

+

y

)

.

i.e.

x

⁢

(

u

+

v

)

⁢

(

v

+

w

+

x

)

+

x

⁢

(

x

⁢

z

+

u

⁢

x

+

u

⁢

y

+

v

⁢

z

+

w

⁢

z

+

u

⁢

w

)

=

(

y

+

z

)

⁢

w

⁢

(

v

+

w

+

x

)

+

(

y

+

z

)

⁢

(

v

⁢

x

+

v

⁢

y

+

w

⁢

y

)

i.e. by (1)

x

⁢

(

x

⁢

z

+

u

⁢

x

+

u

⁢

y

+

v

⁢

z

+

w

⁢

z

+

u

⁢

w

)

=

(

y

+

z

)

⁢

(

v

⁢

x

+

v

⁢

y

+

w

⁢

y

)

i.e. by (3)

x

⁢

(

x

⁢

z

+

u

⁢

y

+

v

⁢

z

+

w

⁢

z

)

=

(

y

+

z

)

⁢

(

v

⁢

y

+

w

⁢

y

)

.

Using (4), we are down to

x

2

⁢

z

+

x

⁢

u

⁢

y

+

u

⁢

w

⁢

y

+

x

⁢

w

⁢

z

=

(

y

+

z

)

⁢

y

⁢

(

v

+

w

)

i.e. by (3)

x

2

⁢

z

+

v

⁢

y

⁢

(

y

+

z

)

+

x

⁢

w

⁢

z

=

(

y

+

z

)

⁢

y

⁢

(

v

+

w

)

i.e.

x

⁢

z

⁢

(

x

+

w

)

=

(

y

+

z

)

⁢

y

⁢

w

.

But in view of (2), this is the same as (4), and
the proof is

complete

.

Remarks:

Ceva’s

theorem

is an easy

consequence

of (4).

Title

proof of Van Aubel’s theorem

Canonical

name

ProofOfVanAubelsTheorem

Date of creation

2013-03-22 13:58:00

Last modified on

2013-03-22 13:58:00

Owner

mathcam (2727)

Last modified by

mathcam (2727)

Numerical id

8

Author

mathcam (2727)

Entry type

Proof

Classification

msc 51N20

Related topic

VanAubelTheorem

Related topic

ProofOfVanAubelTheorem

Generated on Fri Feb 9 22:13:51 2018 by

LaTeXML
quadrant

quadrant

The

x

- and

y

-axes the

x

⁢

y

-plane in four right-angle domains which are called the

quadrants

of the plane.  They are numbered going round the origin in anticlockwise direction so that

{

(

x

,

y

)

⁢

∣

x

>

⁢

0

,

y

>

0

}

is the

first quadrant

,

{

(

x

,

y

)

∣

x

<

0

,

y

>

0

}

the

second quadrant

, and so on.

Naturally, one can speak of the quadrants of the

complex plane

, too.

The lines

y

=

±

x

have as their

slope angles

±

45

∘

,
thus halving the quadrant angles; they are called the

quadrant

bisectors

.  Cf.

angle bisector as locus

.

Title

quadrant

Canonical

name

Quadrant

Date of creation

2014-12-16 22:42:51

Last modified on

2014-12-16 22:42:51

Owner

Mathprof (13753)

Last modified by

pahio (2872)

Numerical id

14

Author

Mathprof (2872)

Entry type

Definition

Classification

msc 51N20

Synonym

quarter-plane

Related topic

RightAngle

Related topic

ConvexAngle

Related topic

Octant

Defines

1st quadrant

Defines

2nd quadrant

Defines

3rd quadrant

Defines

4th quadrant

Generated on Fri Feb 9 22:13:53 2018 by

LaTeXML
similitude of parabolas

similitude of parabolas

Two

parabolas

need not be

congruent

, but they are always

similar

.  Without the definition of parabola by focus and

directrix

, the fact turns out of the simplest equation

y

=

a

⁢

x

2

of parabola.

Let us take two parabolas

y

=

a

⁢

x

2

and

y

=

b

⁢

x

2

which have the origin as common vertex and the

y

-axis as common axis.  Cut the parabolas with the line

y

=

m

⁢

x

through the vertex.  The first parabola gives

a

⁢

x

2

=

m

⁢

x

,

whence the

abscissa

of the other point of intersection is

m

a

; the corresponding ordinate is thus

m

2

a

.  So, this point has the

position vector

u

→

=

(

m

a

m

2

a

)

=

m

a

⁢

(

1

m

)

Similarly, the cutting point of the line and the second parabola has the position vector

v

→

=

(

m

b

m

2

b

)

=

m

b

⁢

(

1

m

)

Accordingly, those position vectors have the

http://planetmath.org/node/848

linear depencence

a

⁢

u

→

=

b

⁢

v

→

for all values of the slope

m

of the cutting line.  This means that both parabolas are

homothetic

with respect to the origin and therefore also similar.

Title

similitude of parabolas

Canonical name

SimilitudeOfParabolas

Date of creation

2013-03-22 18:51:04

Last modified on

2013-03-22 18:51:04

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

8

Author

pahio (2872)

Entry type

Theorem

Classification

msc 51N20

Classification

msc 51N10

Synonym

similarity of parabolas

Related topic

Homothety

Generated on Fri Feb 9 22:14:03 2018 by

LaTeXML
simplest common equation of conics

simplest common equation of conics

In the plane, the locus of the points having the ratio of their

distances

from a certain point (the focus) and from a certain line (the

directrix

) equal to a given constant

ε

, is a

conic section

, which is an

ellipse

, a

parabola

(

http://planetmath.org/ConicSection

) or a

hyperbola

depending on whether

ε

is less than, equal to or greater than 1.

For showing this, we choose the

y

-axis as the directrix and the point

(

q

,

0

)

as the focus.  The locus condition reads then

(

x

-

q

)

2

+

y

2

=

ε

⁢

x

.

This is simplified to

(

1

-

ε

2

)

⁢

x

2

-

2

⁢

q

⁢

x

+

y

2

+

q

2

=

0

.

(1)

If

ε

=

1

,  we obtain the parabola

y

2

=

2

⁢

q

⁢

x

-

q

2

.

In the following, we thus assume that

ε

≠

1

.

Setting

y

:=

0

in (1) we see that the

x

-axis cuts the locus in two points with the

midpoint

of the segment connecting them having the

abscissa

x

0

=

q

1

-

ε

2

.

We take this point as the new origin (replacing

x

by

x

+

x

0

); then the equation (1) changes to

(

1

-

ε

2

)

⁢

x

2

+

y

2

=

ε

2

⁢

q

2

1

-

ε

2

.

(2)

From this we infer that the locus is

1.

in the case

ε

<

1

an ellipse (

http://planetmath.org/Ellipse2

) with the semiaxes

a

=

ε

⁢

q

1

-

ε

2

,

b

=

ε

⁢

q

1

-

ε

2

and with

eccentricity

ε

;

2.

in the case

ε

>

1

a hyperbola (

http://planetmath.org/Hyperbola2

) with semiaxes

a

=

ε

⁢

q

ε

2

-

1

,

b

=

ε

⁢

q

ε

2

-

1

and also now with the eccentricity

ε

.

equation

the origin into a focus of a conic section (and in the cases of ellipse and hyperbola, the abscissa axis through the other focus).  As before, let

q

be the distance of the focus from the corresponding directrix.  Let

r

and

φ

be the

polar coordinates

of an arbitrary point of the conic.  Then the locus condition may be expressed as

r

q

±

r

⁢

cos

⁡

φ

=

ε

.

Solving this equation for the

http://planetmath.org/node/6968

polar radius

r

yields the form

r

=

ε

⁢

q

1

∓

ε

⁢

cos

⁡

φ

(3)

for the common

polar equation

of the conic.  The sign alternative (

∓

) depends on whether the

polar axis

(

φ

=

0

) intersects the directrix or not.

Title

simplest common equation of conics

Canonical name

SimplestCommonEquationOfConics

Date of creation

2015-03-12 8:24:02

Last modified on

2015-03-12 8:24:02

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

11

Author

pahio (2872)

Entry type

Derivation

Classification

msc 51N20

Synonym

common equation of conics

Related topic

ConicSection

Related topic

QuadraticCurves

Related topic

BodyInCentralForceField

Generated on Fri Feb 9 22:14:05 2018 by

LaTeXML
slope

slope

The

slope

of a line in the

x

⁢

y

-plane expresses how great is the change of the

ordinate

y

of the point of the line per a unit-change of the

abscissa

x

of the point; it requires that the line is not vertical.

The slope

m

of the line may be determined by taking the changes of the

coordinates

between two arbitrary points

(

x

1

,

y

1

)

and

(

x

2

,

y

2

)

of the line:

m

=

y

2

-

y

1

x

2

-

x

1

The equation of the line is

y

=

m

⁢

x

+

b

,

where

b

indicates the

intersection

point of the line and the

y

-axis (one speaks of

y-intercept

).

The slope is equal to the

tangent

(

http://planetmath.org/DefinitionsInTrigonometry

) of the

slope angle

of the line.

Two non-vertical lines of the plane are

parallel

if and only if their slopes are equal.

In the previous picture, the blue line given by

3

⁢

x

-

y

+

1

=

0

has slope

3

, whereas the red one given by

2

⁢

x

+

y

+

2

=

0

has slope

-

2

.  Also notice that positive slopes

represent

ascending graphs and negative slopes represent descending graphs.

Title

slope

Canonical name

Slope

Date of creation

2013-03-22 14:48:10

Last modified on

2013-03-22 14:48:10

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

14

Author

pahio (2872)

Entry type

Definition

Classification

msc 51N20

Synonym

angle coefficient (?)

Related topic

Derivative

Related topic

ExampleOfRotationMatrix

Related topic

ParallellismInEuclideanPlane

Related topic

SlopeAngle

Related topic

LineInThePlane

Related topic

DifferenceQuotient

Related topic

DerivationOfWaveEquation

Related topic

IsogonalTrajectory

Related topic

TangentOfHyperbola

Generated on Fri Feb 9 22:14:08 2018 by

LaTeXML