C.C. License

Claus Johansen, Sønderborg, Danmark, 2019-07-16 -> 2020-01-14


Construct a conic by two points and two tangents




Oversigt



Indhold:

0) Preface

1) The task

2) The solution, ellipse and hyperbola separately

3) The solution, ellipse and hyperbola collectively

4) The solution divided into regions

5) The solution at limit values

6) Solutions in the interval 90° -> 270°




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0) Preface

The curves created by conics (circle, ellipse, parabola and hyperbola) have several good properties, when modelling geometri.

It can be practical to define, which conic connets two points, when a specific angle is demanded in these two points.

There is a continous transition between the different conics: circle, ellipse, parabola and hyperbola.

When it's stated here, that you can define a conic by two two points and two tangents, it's a mild exaggeration - the one point must be a top-point, so that the centre-line thereby is known.




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1) The task

An example is shown below. The one point is placed in the centre of the coordinate system and has thereby the coordinates (0, 0) - this point is a top-point and the x-axis is symmetri-axis for the conics that are determined here. The other point is placed in the coordinates (h, w):

Where:
h ~ the height of the points position (=x coordinate).
w ~ the width of the points position, or the half width of the conic at this place (=y coordinate).

Fig.1.1
Fig. 1.1: The task.



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2) The solution, ellipse and hyperbola separately

The formulas were first deduced for each type of conic separatly, and lead to the following solutions:

No Ellipse Hyperbola property
1 eq.2.1   = do Slope of the tangent in the point (h, w)
2 eq.2.3.a   = do Slope of the radius to the point (h, w)
3 eq.2.4.a eq.2.4.b a (semi axis along x-axis)
4 eq.2.5.a eq.2.5.b Dim.less b (semi axis parallel to y-axis)
5 eq.2.6   = do b (semi axis parallel to y-axis)

At a closer inspection, it's easy to see, that ther's only minor differences between the equations in column 2 and 3: "Ellipse" and "Hyperbola" - it's actually only change of sign, that keeps them apart. It's therefore obvious to combine them to one set of equations - that will be done in the next section.




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3) The solution, ellipse and hyperbola collectively

The slope is evaluated first:

eq.2.1   (3.1)

Then the property "d" is evaluated - it determines the type of conic (it can be recognised as part of the expressions in row 2, in the table in the previous section, where "τ" is evaluated):

eq.2.2   (3.2)

Thereafter τ:

eq.2.3   (3.3)

The semi axis along the X-axis:

eq.2.4.a   (3.4)

The dimensionless semi axis parallel to the Y-axis:

eq.2.5   (3.5)

Thereby the semi axis parallel to the Y-axis becomes:

eq.2.6   (3.6)

Some examples of possible solutions are shown in the figure below (there are no solutions in the gray coloured areas):

Fig.3.1
Fig. 3.1: Examples of solutions.

There are a couple reservation in this solution:

These reservation will be treated i § 5.




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4) The solution divided into regions

The regions with uniform solutions are examined systematically below , where θ take values from -90° to +90°.


4.a) -90° < θ < 0°

Fig.4.1
Fig. 4.1: -90° < θ < 0°.

NB: The value for θ of 0° is only valid in this example.

By comparing the semi-axes of the ellipse, a and b, you will find:


4.b) 0° < θ < 26.565°

Fig.4.2
Fig. 4.2: 0° < θ < 26.565°.

NB: The value for θ of 26.565° is only valid in this example, the criterion is:
d < 0 ⇒ ellipse

It will still be relevant, to make a check of the shape of the ellipse:


4.c) 26.565° < θ < 45°

Fig.4.3
Fig. 4.3: 26.565° < θ < 45°.

NB: The value for θ of 26.565° is only valid in this example, the criterion is:
d > 0 ⇒ hyperbel

NB: The value for θ of 45° is also only valid in this example, the criterion is, that θ is smaller than the angle of the line that connects the two points:
θ < atan(w/h).


4.d) 45° ≤ θ ≤ 90°

There shouldn't be any solutions in this interval, but the equations don't give up with a negative square root or similar. The solutions that appears are hyperbolas, but they don't fulfill the criterion to pass through the point (h, w):

Fig.4.4
Fig. 4.4: 45° ≤ θ ≤ 90°

NB: The value for θ of 45° is also only valid in this example, the criterion is, that θ is smaller than the angle of the line that connects the two points:
θ < atan(w/h).




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5) The solution at limit values

There was a couple of reservations to the general solution, mentioned in § 3.

Formally you could argue, that with the right angle, there could be solutions to the problem in the form of straight lines, if either h = 0 or w = 0. And further you could argue, that a straight line just is a degenerated conic - this is mostly curiosities...


5.a) m = 0

At m = 0 there'll be divided by 0 i eq. 3.3 and 3.4, and the set of equations can't be used. The solution is however very simple.

For m = 0 is the answer simply an ellipse, with the two semi-axis h and w:

eq.5.a.1   (5.a.1)
eq.5.a.2   (5.a.2)

With the chosen set of numbers (h, w) = (1, 1) the ellipse becomes a circle when m = 0 and thereby θ = 0. This is however a pure coincidence, the solutions for m = 0 are shown below for the set of numbers (h, w) = (1, 2) and (h, w) = (2, 1):

Fig.5.a.1 Fig.5.a.2
Fig. 5.a.1: m = 0 & (h, w) = (1, 2). Fig. 5.a.2: m = 0 & (h, w) = (2, 1).

5.b) d = 0

When d = 0 leads to τ = 0 in eq. 3.3 and ther will threby by divided by 0 in eq. 3.4, and the set of equations can't be used. The solution is however very simple.

For d = 0 is the answer simply a parabola.

The equation for a lying parabola can be written as:

eq.5.b.1   (5.b.1)

When it's known, that the curve must pass through the point (h, w), it's easy to evaluate the constant c:

eq.5.b.2   (5.b.2)

You can also choose an equation, that more directly gives y as function of x:

eq.5.b.3   (5.b.3)

When it's still known, that the curve must pass through the point (h, w), it's easy to evaluate the constant k:

eq.5.b.4   (5.b.4)
Fig.5.b
Fig. 5.b: The parabola

5.c) a = b

There's per se nothing special by the case a = b - it's just an ellipse with to equal axes, in other words a circle. Here it can be seen as a limit-case in the way, thet it separates the ellipses that are standing, from the lying ones.

Fig.5.c.1 Fig.5.c.2
Fig. 5.c.1: Cirkle for (h, w) = (1, 2). Fig. 5.c.2: Cirkle for (h, w) = (2, 1).



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6) Solutions in the interval 90° -> 270°

The solutions in the interval 90° -> 270° repeats simply the solutions that were found in the interval -90° -> 90°. As it's illustrated below, it doesn't change the analysed conic, if the tangents are rotated 180°:

Fig.6.1 Fig.6.2
Fig. 6.1: Tangents -90° & 0°. Fig. 6.2: Tangents +90° & 180°.



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Oversigt