Proof of Conic Sections
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Transcript of Proof of Conic Sections
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