I take as a definition of a <a href=lemniscate.html>lemniscate</a> the set of points the product of whose distances from (−1, 0) and from (1, 0) is 1.
Squaring both sides gives us
<br>((x+1)<sup>2</sup>+y<sup>2</sup>)((x−1)<sup>2</sup>+y<sup>2</sup>) = 1
<br>(x<sup>2</sup> + 2x + 1 +y<sup>2</sup>)(x<sup>2</sup> − 2x + 1 +y<sup>2</sup>) = 1
<br>x<sup>4</sup> − 2x<sup>2</sup> + y<sup>4</sup> + 2 x<sup>2</sup>y<sup>2</sup> + 2y<sup>2</sup> = 0.
<p>This cure passes thru (√2, 0) and we scale this cure down by a factor of √2 so that it goes thru (1, 0) to match the ‘unit’ lemniscate.
<br>x<sup>4</sup> − x<sup>2</sup> + y<sup>4</sup> + 2x<sup>2</sup>y<sup>2</sup> + y<sup>2</sup> = 0.
<hr>Substituting x for x<sup>2</sup> and y for y<sup>2</sup> we get:
<br>x<sup>2</sup> + 2xy + y<sup>2</sup> − x + y = 0
<p>x+y and x−y are a fine pair of orthogonal coordinates wherein this equation is
<br>(x+y)<sup>2</sup> = −(x−y), a manifest parabola with vertex at 0 and axis thru (x, y) = (1, −1).
(x, y) points (0, 0), (1, 0) and (0, −1) are on the parabola.
<p>(x<sup>2</sup>, y<sup>2</sup>) is on the parabola just in case (x, y) is on the lemniscate!
The map (x, y) ↔ (x<sup>2</sup>, y<sup>2</sup>) of the plane to the plane is not generally interesting geometrically.
This is a special hack.
<p>If r is a real rational then 2/(1+ir) − 1 is a point on the complex plane with rational coefficients.
It is on the unit circle and this formula produces a dense set of (sin t, cos t) pairs on the circle with rational coefficients.
<a href=http://mathworld.wolfram.com/Lemniscate.html>Wolfram</a> gives a parametric form for the lemniscate as:
<br>x = (cos t)/(1 + sin<sup>2</sup> t)
<br>y = (cos t)(sin t)/(1 + sin<sup>2</sup> t)
<br>and we have rational parametric equations for the lemniscate.
<hr>Scheme
<pre>(define i (sqrt -1))
(define (c r) (- (/ 2 (+ 1 (* i r))) 1))
(define (cs r) (let ((y (c r))) (cons (real-part y) (imag-part y))))
(define (w q)(let* ((p (cs q))(sin (cdr p))(cos (car p))(d (/ (+ 1 (* sin sin)))))
   (cons (* cos d) (* cos sin d))))
(define (P x y) (let ((x2 (* x x))(y2 (* y y)))(+ (* x2 x2)(* y2 y2)(* 2 x2 y2) y2 (- x2))))
(define (t r) (let ((p (w r))) (P (car p)(cdr p))))</pre>
For rational r, (c r) is a rational point on the unit circle and (w r) produces a point on the lemniscate with rational coordinates.
Indeed (t r) evaluates the defining polynomial at that point.
(t r) produces exactly 0 for a dozen random rational numbers that I tried.
<p>Here are some rational points on the lemniscate:
<table><tr><td>r<td>(cs r)<td>(w r)<td>(t r)
<tr><td>0<td>(1 . 0)<td>(1 . 0)<td>0
<tr><td>1<td>(0 . 1)<td>(0 . 0)<td>0
<tr><td>−1<td>(0 . −1)<td>(0 . 0)<td>0
<tr><td>2<td>(−3/5 . −4/5)<td>(−15/41 . 12/41)<td>0
<tr><td>−2<td>(−3/5 . 4/5)<td>(−15/41 . −12/41)<td>0
<tr><td>½<td>(3/5 . 4/5)<td>(15/41 . 12/41)<td>0
<tr><td>3<td>(−4/5 . −3/5)<td>(−10/17 . 6/17)<td>0
<tr><td>⅓<td>(4/5 . 3/5)<td>(10/17 . 6/17)<td>0
<tr><td>⅔<td>(5/13 . 12/13)<td>(65/313 . 60/313)<td>0
<tr><td>3/5<td>(8/17 . 15/17)<td>(68/257 . 60/257)<td>0
</table>
<p>Our curve is closed for some value of T near 3.307.
Is this a Lemniscate?
<p>We take p = (x, y) to be the vector position on the curve as a function of t or λ, the arc length.
The curvature is |d<sup>2</sup>p/dλ<sup>2</sup>| = (d<sup>2</sup>p/dt<sup>2</sup>)×(dp/dt)/(dλ/dt)<sup>3</sup>.
(dλ/dt)<sup>2</sup> = (dx/dt)<sup>2</sup> + (dy/dt)<sup>2</sup>.

<hr>Unnecessary work:
<br>Turn by π/4:
Substitute x+y for x and x−y for y:
<br>(x+y)<sup>2</sup> + 2(x<sup>2</sup> − y<sup>2</sup>) + (x−y)<sup>2</sup> − 2(x+y) + 2(x−y) = 0
<br>2x<sup>2</sup> + 2y<sup>2</sup> + 2x<sup>2</sup> − 2y<sup>2</sup> − 4y = 0
<br>x<sup>2</sup> + y = 0.
<p>Back substituting: (x−y)/2 for x and (x+y)/2 for y:
<br>(x−y)<sup>2</sup>/4 + (x+y)/2 = 0
<br>(x−y)<sup>2</sup> + 2(x+y) = 0.
<hr>Dead end:
<br>(x+a)<sup>2</sup> + 2(x+a)(y+b) + (y+b)<sup>2</sup> − 2(x+a) + 2(y+b) = c
<br>x<sup>2</sup> + 2ax + a<sup>2</sup> + 2xy + 2xb + 2ya + 2ab + y<sup>2</sup> + 2yb + b<sup>2</sup> − 2x − 2a + 2y + 2b = c
<br>x<sup>2</sup> + x(2a+2b−2) + 2xy + y(2a+2b+2) + y<sup>2</sup> = c − a<sup>2</sup> − b<sup>2</sup> − 2ab + 2a  − 2b
<br>This requires that 2a+2b−2 = 2a+2b+2 = 0 which is clearly impossible.
It is thus not a centrally symmetric conic—probably a parabola!
<br>(x+y)<sup>2</sup> + 4y = 1
<hr>Back to x<sup>2</sup> + 2xy + y<sup>2</sup> − 2x + 2y = 0
<br>For y=0 we have <br>x<sup>2</sup> − 2x = 0: x=0 or x=2.
<br>For y=1 we have<br>x<sup>2</sup> + 2x + 1 − 2x + 2 = 0
<br>x<sup>2</sup> + 3 = 0 => false
<br>For y=−1 we have<br>x<sup>2</sup> − 2x + 1 − 2x − 2 = 0
<br>x<sup>2</sup> − 4x + 3 = 0 => 

<br>We collect points: (0, 0), (2, 0), (3, −1), (1, −1)