Re: Ellipse equation simplification on Mathematica:

*To*: mathgroup at smc.vnet.net*Subject*: [mg76946] Re: Ellipse equation simplification on Mathematica:*From*: Narasimham <mathma18 at hotmail.com>*Date*: Wed, 30 May 2007 05:29:28 -0400 (EDT)*References*: <f2emof$35h$1@smc.vnet.net><200705280500.BAA15766@smc.vnet.net>

On May 29, 2:19 pm, Andrzej Kozlowski <a... at mimuw.edu.pl> wrote: > On 29 May 2007, at 12:15, Andrzej Kozlowski wrote: > > > > > > > On 29 May 2007, at 10:53, Andrzej Kozlowski wrote: > > > >> On 28 May 2007, at 14:00, Narasimham wrote: > > >>> On May 19, 1:54 pm, Andrzej Kozlowski <a... at mimuw.edu.pl> wrote: > >>>> On 18 May 2007, at 19:06, Narasimham wrote: > > >>>>> Reference is made to: > > >>>>>http://groups.google.co.in/group/geometry.puzzles/browse_thread/ > >>>>> threa... > > >>>>> Constant[c,d th,ph] ; > > >>>>> (* th, ph are spherical cords of tip of tube *) ; > > >>>>> cp = Cos[ph] ; sp = Sin[ph] ; cth = Cos[th] ; sth = Sin[th] ; > > >>>>> (* earlier typo corrected *) > > >>>>> d1 = Sqrt[(x + d cp cth + c )^2 + ( y + d cp sth )^2 + (d sp)^2 ] > > >>>>> d2 =Sqrt[(x - d cp cth - c )^2 + ( y - d cp sth )^2 + (d sp)^2 ] > > >>>>> FullSimplify[ d1 + d2 + 2 d - 2 a == 0] ; > > >>>>> When d = 0, algebraic/trigonometric simplification brings about > >>>>> common ellipse form: > > >>>>> (x/a)^2 + y^2/(a^2-c^2) = 1 > > >>>>> Request help for bringing to standard form involving constants > >>>>> a,c > >>>>> and the new tube length constant d. > > >>>>> Regards, > >>>>> Narasimham > > >>>> I don't think such a form exists. Consider the following. > > >>>> id1 = {d1^2 - ((x + d*cp*cth + c)^2 + (y + d*cp*sth)^2 + (d*sp)^2), > >>>> d2^2 - ((x - d*cp*cth - c)^2 + (y - d*cp*sth)^2 + (d*sp)^2), > >>>> sp^2 + cp^2 - 1, sth^2 + cth^2 - 1}; > > >>>> id = Prepend[id1, d1 + d2 + 2 d - 2 a]; > > >>>> Now consdier first the case of the ellipse: > > >>>> d = 0; > > >>>> gr = GroebnerBasis[id, {x, y, a, c}, {cp, sp, cth, sth, d1, d2}, > >>>> MonomialOrder -> EliminationOrder] > >>>> {-a^4 + c^2 a^2 + x^2 a^2 + y^2 a^2 - c^2 x^2} > > >>>> This tells us that > > >>>> First[%] == 0 > >>>> -a^4 + c^2*a^2 + x^2*a^2 + y^2*a^2 - c^2*x^2 == 0 > > >>>> is the equation of the ellipse, and this can be easily brought to > >>>> standard form by hand. But now consider your "general" case: > > >>>> Clear[d] > >>>> gr = GroebnerBasis[id, {x, y, a, c, d}, {cp, sp, cth, sth, d1, > >>>> d2}, > >>>> MonomialOrder -> EliminationOrder] > >>>> {} > > >>>> This means that elimination cannot be performed and no "standard > >>>> form" > >>>> of the kind you had in mind exists. Unless of course there is a bug > >>>> in GroebnerBasis (v. unlikely) or I have misunderstood what you had > >>>> in mind. > > >>>> Andrzej Kozlowski > > >>> I checked for case of tube parallel to x- or y-axis produces > >>> ellipses > >>> and suspected validity even in 3-D general case. > > >>> Narasimham > > >> OK., now I see that I misundertood you and you wrote that cd,th, > >> ph (and presumably a) are supposed to be constants, so you do not > >> wish to eliminate them. But now one can easily prove that what you > >> get is not, in general, an ellipse. In this situation Groebner > >> basis works and you can obtain a rather horrible quartic equation > >> of your surface: > > >> id1 = {d1^2 - ((x + d*cp*cth + c)^2 + (y + d*cp*sth)^2 + (d*sp)^2), > >> d2^2 - ((x - d*cp*cth - c)^2 + (y - d*cp*sth)^2 + (d*sp)^2), > >> sp^2 + cp^2 - 1, sth^2 + cth^2 - 1}; > > >> id = Prepend[id1, d1 + d2 + 2*d - 2*a]; > > >> v = GroebnerBasis[id, {x, y, a, c, cp, cth}, {sth, sp, d1, d2}, > >> MonomialOrder -> EliminationOrder][[1]]; > > >> First[v] == 0 > > >> is the equation (I prefer not to include the output here). > > >> Looking at v see that the non zero coefficients are only the free > >> coefficient, the coefficients of x^2, y^2, x^2 y^2, x^4 and y^4. > >> So only in some cases you will get a quadratic (for example when > >> the quartic happens to be a perfect square as in the case d=0, or > >> when the free coefficient vanishes, as in the trivial case a=d, > >> or when the coefficients of 4-degree terms vanish). One can work > >> out all the cases when gets a quadratic but it is also easy to > >> find those when one does not. For example, taking both th and ph > >> to be 60 degrees (so that cth and cph are both 1/2) we get: > > >> v /. {cp -> 1/2, cth -> 1/2, d -> 4, a -> 2, c -> 1} > > >> y^4 - 48 x^2 y^2 + 120 y^2 + 3600 > > >> This is certianly is not the equation of an ellipse. > > >> Andrzej Kozlowski > > > Sorry; that last example was a bad one, because with these > > parameters (d>a) the original equations do not have solutions. The > > point is that the equations that we get after elimination will have > > more solutions than the ones we start with, but all the solutions > > of the original ones will satisfy the new ones. So to see that we > > do not normally get an ellipse we need a different choice of > > parameters, with a>d. So take > > instead > > > w = First[v ]/. {cp -> 1/2, cth -> 1/2, sp -> Sqrt[3]/2, sth -> Sqrt > > [3]/2, d -> 1, > > a -> 3, c -> 1} > > > (1521*x^4)/256 + (2229*y^2*x^2)/128 - (117*x^2)/4 + (3721*y^4)/256 > > - (183*y^2)/4 + 36 > > > Now, this is clearly not the equation of an ellipse. We can see now > > the relationship between this and your original equation. With the > > above values of the parameters your equation takes the form: > > > eq = d1 + d2 + 2*d - 2*a == 0 /. {cp -> 1/2, cth -> 1/2, sp -> Sqrt > > [3]/2, > > sth -> Sqrt[3]/2, d -> 1, a -> 3, > > c -> 1} > > Sqrt[(x - 5/4)^2 + (y - Sqrt[3]/4)^2 + 3/4] + > > Sqrt[(x + 5/4)^2 + (y + Sqrt[3]/4)^2 + 3/4] - 4 == 0 > > > So now look at the graph: > > > gr1=ContourPlot[ > > Sqrt[(x - 5/4)^2 + (y - Sqrt[3]/4)^2 + 3/4] + > > Sqrt[(x + 5/4)^2 + (y + Sqrt[3]/4)^2 + 3/4] - 4 == 0, {x, -5, > > 5}, {y, -5, > > 5}] > > > Looks like a nice ellipse, right? Unfortunately it is only a part > > of the graph of > > > w =First[v] > > > gr2 = ContourPlot[Evaluate[w == 0], {x, -5, 5}, {y, -5, 5}] > > > The picture is not quite convincing, because the latter contour > > plot is not very accurate but when you see them together: > > > Show[gr1, gr2] > > > the relationship becomes obvious. In any case, what you get a > > quartic curve that looks of course like an ellipse but isn't. > > > Andrzej Kozlowski > > Sorry, a small correction is needed. I wrote: > > > w=First[v] > > > gr2 = ContourPlot[Evaluate[w == 0], {x, -5, 5}, {y, -5, 5}] > > Ignore the first line (w=First[v]) in the above since it will prevent > the plot from working. Just evaluate the second line with the value > of w defined earlier. > > Andrzej Kozlowski Thanks Andrzej so much for all the cumbersome & involved verifications. Ok,are we now in a situation to find out what exhaustively all situations/conditions give ellipses? Should coefficients of third and fourth order term vanish,or should it have repeated roots etc.? At least don't we get ellipses for tube center when th = 0,? (inclusive when the tube is parallel or perpendicular to x- axis, th = 0,ph = 0,Pi/2? For this case new a,b need to be defined in terms of c,a and d). Regards, Narasimham

**Follow-Ups**:**Re: Re: Ellipse equation simplification on Mathematica:***From:*Andrzej Kozlowski <akoz@mimuw.edu.pl>

**References**:**Re: Ellipse equation simplification on Mathematica:***From:*Narasimham <mathma18@hotmail.com>

**Re: Iterate through a list**

**Re: Iterate through a list**

**Re: Re: Ellipse equation simplification on Mathematica:**

**Re: Re: Ellipse equation simplification on Mathematica:**