In the Cartesian plane is one defines a parabola to be the set of all points equal distant from a fixed line (directrix) and fixed point (focus) one gets there are two standard forms: (x-h)^2=4p(y-k) or (y-k)^2=4p(x-h). In the Poincare half plane we of course can consider both types of geodesic for the 'fixed line.' If one does this and applies that definition can one get standard forms? I tried for a year and got nowhere.
Dear Daniel;
I wish this link my help:
https://arxiv.org/pdf/1603.09285.pdf
Regards
Dear Daniel;
I wish this link my help:
https://arxiv.org/pdf/1603.09285.pdf
Regards
Dear Sir,
For an ellipse, with focus points F1 and F2, I find
cosh(d1) = 1 + ((x-x1)² + (y-y1)²)/(yy1)
cosh(d2) = 1 + ((x-x2)²+(y-y2)²)/(yy2)
To get rid of the cosh, we get
cosh(d1) = cos(id1)
It is an ellipse when d1+d2 = dtot. Use this equation to calculate cosh(d2)
cosh(d2) = cos(id2) = cos(idtot - d1) = cos(idtot)*cos(id1) + sin(idtot)*sin(id1)
= cos(idtot)*cos(id1) + sin(idtot) * sqrt(1 - cos(id1)^2)
Now we replace cos(id1) with equation1 (eq1) and cos(id2) with eq2. So, then we get
eq2 = cos(idtot)*eq1 + sin(idtot)*sqrt(1 -eq1²)
To get rid of the square root, we move everything else to the left hand side
(eq2 - cos(idtot)*eq1)^2/(sin(idtot)^2) = 1 -eq1^2 ;
In the form above, you have to fill in the formulae for eq1 and eq2, do the math, and then you get the equation of the ellipse.
In my opinion, the ellipse becomes a parabole if one of the points F1 or F2 is a point at infinity, meaning y1 = 0 or x1 = 0. I think you need to do some extra math to get rid of the singularity, but in essense it should be similar.
Dear Daniel Replogle,
I am cordially apologizing for my former answer, which is obviously not any answer to your question! I didn't take into account that you are asking about the hiperbolic geometry on the half-plane. So please, take the answer as non-existing!
Best
PS. If possible, I shall remove it.
Dear Daniel Replogle,
So, the inappropriate aswer is removed! Now I am presenting an explicite solution to your problem in the case of directrix being a half-circle, with the focus outside the half-disk (see the attachments).
Regards
Additional remarks:
1. Correction: The last term in (6) should be $y_F $ instead of $v^{\star}(x,y)$
2. Remark: After this, the equation for parabola simplifies to the following one:
$y_F\cdot [ (x^2+y^2+r^2)^2 - 4\cdot x^2\cdot r^2 ] = r^2\cdot [( x-x_F)^2 +y^2+{y_F}^2 ]^2
\leqno(7)$
which is of 4th total degree, the same (degree:) as for the case of $x_F=0$ presented in the paper by P. Chao & J. Rosenberg referred to by Jamal Salah in his answer above. Frankly, the paper helped much in getting the right way via the staight line $p^{\star}$.
Thanks to all RGaters for their patience in reading this rain of mistakes.
Regards
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