Mathematica Asked on August 1, 2021
FullSimplify[( Sqrt[(x + h)^2 + y^2] - a) - e (Sqrt[(x - h)^2 + y^2] - b) == 0]
How to remove the radicals to obtain an algebraic expression of fourth order using Simplify, without squaring by hand, transposing etc. ? Variables are $(x,y)$ and constants $(a,b,h,e)$.
And how for the simpler case of the ellipse of second degree? ( when $c$ is constant):
Sqrt[(x-c)^2+y^2]+ Sqrt[(x+c)^2+y^2]== 2 a
I have plotted special cases for verification, but need an algebraic curve formula.
Thanks for assistance.
Use Eliminate
eq = 0 == (Sqrt[(x + h)^2 + y^2] - a) -
e (Sqrt[(x - h)^2 + y^2] - b) /. (-h + x)^2 + y^2 ->
aa /. (h + x)^2 + y^2 -> bb
(* 0 == -a + Sqrt[bb] - (Sqrt[aa] - b) e *)
eli1 = Eliminate[{eq, (-h + x)^2 + y^2 == aa, (h + x)^2 + y^2 ==
bb}, {aa, bb}]
(* a^4 - 4 a^3 b e +
a^2 (6 b^2 e^2 - 2 h^2 - 2 e^2 h^2 - 4 h x + 4 e^2 h x - 2 x^2 -
2 e^2 x^2 - 2 y^2 - 2 e^2 y^2) +
a b e (-4 b^2 e^2 + 4 h^2 + 4 e^2 h^2 + 8 h x - 8 e^2 h x + 4 x^2 +
4 e^2 x^2 + 4 y^2 + 4 e^2 y^2) == -b^4 e^4 + 2 b^2 e^2 h^2 +
2 b^2 e^4 h^2 - h^4 + 2 e^2 h^4 - e^4 h^4 + 4 b^2 e^2 h x -
4 b^2 e^4 h x - 4 h^3 x + 4 e^4 h^3 x + 2 b^2 e^2 x^2 +
2 b^2 e^4 x^2 - 6 h^2 x^2 - 4 e^2 h^2 x^2 - 6 e^4 h^2 x^2 -
4 h x^3 + 4 e^4 h x^3 - x^4 + 2 e^2 x^4 - e^4 x^4 + 2 b^2 e^2 y^2 +
2 b^2 e^4 y^2 - 2 h^2 y^2 + 4 e^2 h^2 y^2 - 2 e^4 h^2 y^2 -
4 h x y^2 + 4 e^4 h x y^2 - 2 x^2 y^2 + 4 e^2 x^2 y^2 -
2 e^4 x^2 y^2 - y^4 + 2 e^2 y^4 - e^4 y^4 *)
Correct answer by Akku14 on August 1, 2021
I don't think there is an easy way using Simplify
.
I have hacked together a solution that works at least for your examples.
sqrtEqExpand[eqExpr_] := Module[{eq = eqExpr, expr, sqrts, rest},
While[Not@FreeQ[eq, Power[_, 1/2]],
expr = Collect[#, Power[_, 1/2]] &@(ExpandAll@Subtract @@ eq);
sqrts = Cases[expr, ___ Power[_, 1/2] | Power[_, 1/2]];
rest = Complement[List @@ expr, sqrts];
eq = ExpandAll[
First[sqrts]^2 == (-Plus @@ Rest[sqrts] - Plus @@ rest)^2];];
eq]
Examples are
eq1 = (Sqrt[(x + h)^2 + y^2] - a) - e (Sqrt[(x - h)^2 + y^2] - b) == 0;
sqrtEqExpand[eq1]
(* 4 a^2 h^2 - 8 a b e h^2 + 4 b^2 e^2 h^2 + 8 a^2 h x - 16 a b e h x +
8 b^2 e^2 h x + 4 a^2 x^2 - 8 a b e x^2 + 4 b^2 e^2 x^2 +
4 a^2 y^2 - 8 a b e y^2 + 4 b^2 e^2 y^2 ==
a^4 - 4 a^3 b e + 6 a^2 b^2 e^2 - 4 a b^3 e^3 + b^4 e^4 +
2 a^2 h^2 - 4 a b e h^2 - 2 a^2 e^2 h^2 + 2 b^2 e^2 h^2 +
4 a b e^3 h^2 - 2 b^2 e^4 h^2 + h^4 - 2 e^2 h^4 + e^4 h^4 +
4 a^2 h x - 8 a b e h x + 4 a^2 e^2 h x + 4 b^2 e^2 h x -
8 a b e^3 h x + 4 b^2 e^4 h x + 4 h^3 x - 4 e^4 h^3 x + 2 a^2 x^2 -
4 a b e x^2 - 2 a^2 e^2 x^2 + 2 b^2 e^2 x^2 + 4 a b e^3 x^2 -
2 b^2 e^4 x^2 + 6 h^2 x^2 + 4 e^2 h^2 x^2 + 6 e^4 h^2 x^2 +
4 h x^3 - 4 e^4 h x^3 + x^4 - 2 e^2 x^4 + e^4 x^4 + 2 a^2 y^2 -
4 a b e y^2 - 2 a^2 e^2 y^2 + 2 b^2 e^2 y^2 + 4 a b e^3 y^2 -
2 b^2 e^4 y^2 + 2 h^2 y^2 - 4 e^2 h^2 y^2 + 2 e^4 h^2 y^2 +
4 h x y^2 - 4 e^4 h x y^2 + 2 x^2 y^2 - 4 e^2 x^2 y^2 +
2 e^4 x^2 y^2 + y^4 - 2 e^2 y^4 + e^4 y^4 *)
eq2 = Sqrt[(c - x)^2 + y^2] - Sqrt[(c + x)^2 + y^2] == 2 a;
sqrtEqExpand[eq2]
(* 16 a^2 c^2 + 32 a^2 c x + 16 a^2 x^2 + 16 a^2 y^2 ==
16 a^4 + 32 a^2 c x + 16 c^2 x^2 *)
Answered by Natas on August 1, 2021
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