Dual curve of the lemniscate of Bernoulli?

Question

Does anybody happen to know the homogeneous equation $F(u,v,w) = 0$ for the algebraic curve dual to the lemniscate of Bernoullli: $$ F(x,y,x) = (x^2 + y^2)^2 - A (x^2 -y^2)z^2 = 0?$$

I looked at the equation of the tangent $(\partial F/\partial x, \partial F/\partial y, \partial F/\partial z) =: (u,v,w)$ i.e.:

$$ u = 4x(x^2+y^2) - 2Axz^2 \\ v = 4y(x^2+y^2) + 2Ayz^2 \\ w = 2Az(y^2-x^2)$$

But I wouldn't know how to guess an equation between $u, v$ and $w$?

Is there somebody that can use Mathematica to Eliminate the solution?

Answer 1

In a trial version of Mathematica I used the Eliminate function like this:
Eliminate[ {(x^2 + y^2) ^2 == A*(x^2-y^2) * z^2, u== 2*x*(x^2 + y^2) - A*x * z^2, v == 2*y*(x^2 + y^2) + A*y * z^2, w == A*z *(y^2 - x^2)}, {x,y,z}].

And the result is: $$ F^d = A^3(u^2 - v^2)^3 + A^2(15u^4 +78u^2v^2 +15v^4)w^2 + A(48u^2 - 48 v^2) w^4 - 64w^6 = 0 $$

The lemniscate has degree 4 and 3 ordinary nodes. The dual curve should then have degree $4*3 - 2* 3 =6$ which seems to hold.

I checked the result using Mathematica again and it turns out to be correct.
With the help of several functions:

• dualLemnVec[{u_, v_, w_}, A_] := A^3 *(u^2 - v^2 )^3 + A^2 *(15 ( u^2 + v^2)^2 + 48 u^2 v^2 )w^2 + A 48( u^2 - v^2)w^4 - 64 w^6
• lemn[{x_,y_,z_}]:=(x^2+y^2)^2 - A*(x^2 -y^2)*z^2
• paramLemn[t_]:={Sqrt[A](t + t^3),Sqrt[A]( t-t^3), 1+ t^4}
• Simplify[lemn[paramLemn[t]]] = 0 (doublecheck)
• diffparamLemn[t_] := {Sqrt[A](1 +3 t^2), Sqrt[A]*(1 - 3*t^2), 4* t^3}
• paramDualLemn[t_]:=Cross[paramLemn[t],diffparamLemn[t]]
• Simplify[dualLemnVec[paramDualLemn[t], A]] = 0 (yes!)

Also Mathematica calculated in a fraction of a second the dual of the dual of the lemniscate and it turned out to be the lemniscate again.

D[dualLemnVec[{u,v,w},A],u] = 6 A^3 u (u^2-v^2)^2+A^2 (96 u v^2+60 u (u^2+v^2)) w^2+96 A u w^4.
Likewise the partial derivates w.r.t. v and w.

Eliminate[{A^2 (-15 u^4 - 78 u^2 v^2 - 15 v^4) w^2 + A (-48 u^2 + 48 v^2) w^4 + 64 w^6 == A^3 * ( u^6 - 3 u^4 v^2 + 3 u^2 v^4 - v^6),
x == 6 *A^3 *u (u^2 - v^2)^2 + A^2 (96 u v^2 + 60 u (u^2 + v^2)) w^2 + A 96 u w^4,
y == -6 *A^3 *v (u^2 - v^2)^2 +A^2 * (96 u^2 v + 60 v (u^2 + v^2)) w^2 - A* 96 v w^4, z == 2*A^2* (48 u^2 v^2 + 15 (u^2 + v^2)^2) w + 192 A(u^2 - v^2) w^3 - 384 w^5}, {u, v, w}]

Answer surprisingly or not: (x^2 + y^2) ^2 == A*(x^2-y^2) * z^2

Some easy points of the dual lemniscate:

• $(1,\pm 1, 0)$: two asymptotes $x=\pm y$, their points of infinity are cusps;
• $(\pm \sqrt{1/A}, 0, 1)$: two points on the x-axis;
• $(\pm i\sqrt{8/A}, 0, 1)$: two imaginary points on the x-axis;
• $(0,\pm \sqrt{8/A}, 1)$: two nodes on the y-axis;
• $(0, \pm i\sqrt{1/A}, 1)$: two imaginary points on the y-axis;