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Hyperbola transformed to a canonical form

ValeryOchkov
24-Ruby IV

Hyperbola transformed to a canonical form

I bring the curve of the second order (hyperbola - red curve) to the canonical form. Why don't my starting blue points end up on a hyperbola transformed to a canonical form. Help me please!

Mathcad 15 in attach.

ValeryOchkov_0-1664887070561.png

 

1 ACCEPTED SOLUTION

Accepted Solutions

Instead of changing the value of the angle or using (pi/2-phi) instead of phi you may also just exchange sin and cos in your two rotation equations.

Werner_E_0-1664900500277.png

I also wondered why you used the "root" function to calculate phi instead of a direct calculation

Werner_E_1-1664900563160.png

 

This time the sheet is attached 😉

View solution in original post

17 REPLIES 17

1) You used X instead of X1 in yor second plot

 

2) The center of the original hyperbola is not in the origin, so you have to translate X and Y by x.o and y0 before you rotate them

Werner_E_0-1664889549061.png

3) You angle phi is wrong. Here you see the original hyperbola (matrix H) translated by (x.0; y.0) and rotated by your angle phi

Werner_E_1-1664889630917.pngWerner_E_2-1664889646109.png

 

The correct angle to rotate is not 42.659° but rather 90° - 42.659° = 47.341°.

 

Using all these corrections we get

Werner_E_4-1664889891888.png

 

Thanks, Werner!

Send please the sheet!


@ValeryOchkov wrote:

Thanks, Werner!

Send please the sheet!


Sorry, I did not save and keep the modified sheet.

But only very few changes are necessary and they can easily be taken from the pictures, I think.

  • Simply add the angle to 90°,
  • add the displacement at the calculation of X1 and Y1 as shown
  • and use X1 instead of X in the plot.

That is all that is necessary.

Instead of changing the value of the angle or using (pi/2-phi) instead of phi you may also just exchange sin and cos in your two rotation equations.

Werner_E_0-1664900500277.png

I also wondered why you used the "root" function to calculate phi instead of a direct calculation

Werner_E_1-1664900563160.png

 

This time the sheet is attached 😉

Some additional remarks:

You could get the coefficients for your implicit equation more efficient by using the symbolic "coeff" command so you can avoid copy&paste:

Werner_E_0-1664904188041.png

 

And while I think that implicitplot2D() is a most valuable tool, you don't need it to plot hyperbolas as you could use a parameter equation

Werner_E_1-1664904261943.png

 

Dummy is x^3.

 


@ValeryOchkov wrote:

Dummy is x^3.

 


No, "dummy" is the coefficient of x^5 which would correspond to (non-existent) x^2*y in the original expression.

Yes! With dynamic: sh(t), ch(t)...

Fig-1-XY.pngFig-2-XYPlot-1.pngFig-3-XYPlot-2.pngFig-4-XYPlot-Root.pngFig-5-SLAE.pngFig-6-Wiki-2.pngFig-7-SLAE-Fi.pngFig-8-Roots.pngFig-9-Canonic.png

ValeryOchkov_1-1664950043625.png

 

ValeryOchkov_0-1664949967196.png

 

Origin seven (five) points (the Asteroid position in time), our blue planet, two branches of hyperbola, two focus, two asymptotes, two directrices and... new method  of hyperbola transforming to a canonical form. Or it is one old method?  

ValeryOchkov_1-1664906395585.png

 

 

Sorry, I don't understand your question.

Is it one new method of transforming one curve of 2-d order in canonical form. With using one Math package?

I guess its not really new but rather a bit different from the usual way of using matrices, eigenvalues, etc. But the basic math is quite similar, I suppose.

Many students are horrified by words eigenvalues, etc. And so - "not aesthetically pleasing, but cheap, reliable and practical" using the numerical solution of systems of equations.

SMath

ValeryOchkov_0-1664909322087.png

(67) Лёлик, но это же не эстетично! Зато дёшего, надёжно и практично! - YouTube

 


@ValeryOchkov wrote:

Many students are horrified by words eigenvalues, etc. And so - "not aesthetically pleasing, but cheap, reliable and practical" using the numerical solution of systems of equations.


In other words .. KISS 😄

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