PROOF of CONIC SECTIONS

Equation of Conical surface shown below is , (easily found

by elementary geometry).

Take a plane never intersecting y axis. That is, .

Side view of the plane is,

Where

.

Now, equation of the cone is constrained by cutting plane, on which 2D equations

are generated; coordinates must transformed into new coordinates

shown below to project x,y,z values on to cutting plane.

Note that u,v axes are intersecting z axis (lying on y=0 plane)

Proof of transformation:

Take an arbitrary point called p.

With elementary geometry using figure above following relations can be

obtained.

( )

( )

Equation of conical surface becomes,

( ) ( )

And equation of the plane becomes

Now equation of intersection curve is just equation of the cone with substituting

( ) ( )

Now we can work on plane. For 2D coordinate system, curves can be shown

by single equation, unlike, for 3D space shown by 2 equations. Therefore, we

disregard equation anymore.

Arranging last equation,

( )

( ) ( )

General form:

Note that, all equations must be symmetric with respect to u axis by inspection

since cone is a symmetrical solid. Therefore, for all equations ( ) ( )

Special Cases:

Parabola

If coefficient of is zero we have

( )

Which is equation of parabola in the form of

(coefficient of )

Multiplying all by ,

( )( )

And using trigonometric identities we have

( ) ( )

We have two cases

If ( ) ,

If ( ) ,

First case is redundant due to symmetry.

Result: If a cone is cut parallel to the lateral surface, parabolic profile is

obtained.

Circle

If coefficient of is zero and coefficient of is -1: circle is obtained,

.

( ) ( )

cannot be zero to form a cone, n cannot be zero to not intersect origin

insead of cones. Therefore only condition is

Also we should guarantee that

since

Expression becomes -1 automatically.

Result: If a cone is cut perpendicular to its axis, circle is obtained.

Hyperbola

Let us take

, now life becomes easier since plane of intersection becomes

x=C plane. And our 2D variables become y and z.

Which is equation of a hyperbola in the form:

.

Result: If a cone is cut parallel to its axis hyperbolic profile is obtained.

Ellipse:

Use similar form, because operations are going to be messy.

Make following manipulations

( )

( )

( ) ( √ )

Now A>0 to make signs of coefficients of u and y the same, also;

Solving first restriction,

(

)

(

)

It is possible for

Result: If intersection is bounded everywhere, ellipse is obtained.

Note that circle,bounded everywhere, is a special case of ellipse.

Other cases:

It is known that at that conditions

(

)

Recall:

( ) ( )

Coefficient of :

( )

* (

) +

[

( )

]

Use restriction,

(

)

| |

( )

( )

( )

Hence,

[

( )

]

Coefficient is positive.

Coefficient of :

No idea since sign of n is unknown.

Constant term:

is always positive.

Use simpler form

Again completing square

(√

√ )

Multiply all by 4a

( )

( )

Determination of sign of is important.

( ) ( )( )

[( ) ( )( )]

[( ) ( )( )]

[( ) ( )]

all squares so it is positive.

We have hyperbola of the form

( )

Substitute all values following formulas are obtained.

Formulations:

Equation of the cone:

Equation of the plane:

Cutting Angle Shape Formula

Circle

Ellipse

( )

Parabola

Hyperbola

( )

Shape Constants

Circle | |

Ellipse | |

, | |

√

√

Parabola

Hyperbola | |

( ) , | |

√

√ ( ) ,

Boray