{
  "index": "1941-B-1",
  "type": "ANA",
  "tag": [
    "ANA",
    "GEO"
  ],
  "difficulty": "",
  "question": "8. A particle \\( (x, y) \\) moves so that its angular velocities about \\( (1,0) \\) and \\( (-1,0) \\) are equal in magnitude but opposite in sign. Prove that\n\\[\ny\\left(x^{2}+y^{2}+1\\right) d x=x\\left(x^{2}+y^{2}-1\\right) d y\n\\]\nand verify that this is the differential equation of the family of rectangular hyperbolas passing through \\( (1,0) \\) and \\( (-1,0) \\) and having the origin as center.",
  "solution": "Solution. The angular velocity about the origin of a point moving according to the parametric equations\n\\[\nx=x(t), \\quad y=y(t)\n\\]\nis\n\\[\n\\frac{x \\dot{y}-y \\dot{x}}{x^{2}+y^{2}}\n\\]\nwhere the dots indicate differentiation with respect to \\( t \\). Translating the center of reference, first to \\( (1,0) \\) and then to \\( (-1,0) \\), we can express the condition as\n\\[\n\\frac{(x-1) \\dot{y}-y \\dot{x}}{(x-1)^{2}+y^{2}}+\\frac{(x+1) \\dot{y}-y \\dot{x}}{(x+1)^{2}+y^{2}}=0 .\n\\]\n\nFrom this we get\n\\[\n\\begin{array}{c}\n\\quad(x \\dot{y}-y \\dot{x}-\\dot{y})\\left(x^{2}+y^{2}+1+2 x\\right) \\\\\n+(x \\dot{y}-y \\dot{x}+\\dot{y})\\left(x^{2}+y^{2}+1-2 x\\right)=0,\n\\end{array}\n\\]\nwhich simplifies to\n\\[\nx\\left(x^{2}+y^{2}-1\\right) \\dot{y}=y\\left(x^{2}+y^{2}+1\\right) \\dot{x}\n\\]\n\nThis is equivalent to the required differential equation.\nA central rectangular hyperbola has an equation of the form\n\\[\nA\\left(x^{2}-y^{2}\\right)+B x y=1\n\\]\n\nIt passes through \\( (-1,0) \\) and \\( (1,0) \\) if and only if \\( A=1 \\). So the suggested family is given by\n\\[\nx^{2}-y^{2}+B x y=1\n\\]\n\nDifferentiation with respect to \\( \\boldsymbol{t} \\) yields\n\\[\n2 x \\dot{x}-2 y \\dot{y}+B(x \\dot{y}+\\dot{x} y)=0 .\n\\]\n\nEliminating \\( B \\) between (2) and (3) we get\n\\[\nx y(2 x \\dot{x}-2 y \\dot{y})+(x \\dot{y}+\\dot{x} y)\\left(1-x^{2}+y^{2}\\right)=0 .\n\\]\n\nThis is the differential equation of the suggested family of hyperbolas and it simplifies to (1) as required.",
  "vars": [
    "x",
    "y",
    "t"
  ],
  "params": [
    "A",
    "B"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "abscissa",
        "y": "ordinate",
        "t": "timeline",
        "A": "parama",
        "B": "paramb"
      },
      "question": "8. A particle \\( (abscissa, ordinate) \\) moves so that its angular velocities about \\( (1,0) \\) and \\( (-1,0) \\) are equal in magnitude but opposite in sign. Prove that\n\\[\nordinate\\left(abscissa^{2}+ordinate^{2}+1\\right) d\\,abscissa = abscissa\\left(abscissa^{2}+ordinate^{2}-1\\right) d\\,ordinate\n\\]\nand verify that this is the differential equation of the family of rectangular hyperbolas passing through \\( (1,0) \\) and \\( (-1,0) \\) and having the origin as center.",
      "solution": "Solution. The angular velocity about the origin of a point moving according to the parametric equations\n\\[\nabscissa = abscissa(\\,timeline), \\quad ordinate = ordinate(\\,timeline)\n\\]\nis\n\\[\n\\frac{abscissa \\, \\dot{ordinate} - ordinate \\, \\dot{abscissa}}{abscissa^{2}+ordinate^{2}}\n\\]\nwhere the dots indicate differentiation with respect to \\( timeline \\). Translating the center of reference, first to \\( (1,0) \\) and then to \\( (-1,0) \\), we can express the condition as\n\\[\n\\frac{(abscissa-1) \\, \\dot{ordinate} - ordinate \\, \\dot{abscissa}}{(abscissa-1)^{2}+ordinate^{2}} + \\frac{(abscissa+1) \\, \\dot{ordinate} - ordinate \\, \\dot{abscissa}}{(abscissa+1)^{2}+ordinate^{2}} = 0 .\n\\]\n\nFrom this we get\n\\[\n\\begin{array}{c}\n\\quad(abscissa \\, \\dot{ordinate} - ordinate \\, \\dot{abscissa} - \\dot{ordinate})\\left(abscissa^{2}+ordinate^{2}+1+2abscissa\\right) \\\\\n + (abscissa \\, \\dot{ordinate} - ordinate \\, \\dot{abscissa} + \\dot{ordinate})\\left(abscissa^{2}+ordinate^{2}+1-2abscissa\\right) = 0 ,\n\\end{array}\n\\]\nwhich simplifies to\n\\[\nabscissa\\left(abscissa^{2}+ordinate^{2}-1\\right) \\dot{ordinate} = ordinate\\left(abscissa^{2}+ordinate^{2}+1\\right) \\dot{abscissa}\n\\]\n\nThis is equivalent to the required differential equation.\nA central rectangular hyperbola has an equation of the form\n\\[\nparama\\left(abscissa^{2}-ordinate^{2}\\right) + paramb \\, abscissa \\, ordinate = 1\n\\]\n\nIt passes through \\( (-1,0) \\) and \\( (1,0) \\) if and only if \\( parama = 1 \\). So the suggested family is given by\n\\[\nabscissa^{2}-ordinate^{2} + paramb \\, abscissa \\, ordinate = 1\n\\]\n\nDifferentiation with respect to \\( \\boldsymbol{timeline} \\) yields\n\\[\n2abscissa\\,\\dot{abscissa} - 2ordinate\\,\\dot{ordinate} + paramb\\,(abscissa\\,\\dot{ordinate} + \\dot{abscissa}\\,ordinate) = 0 .\n\\]\n\nEliminating \\( paramb \\) between (2) and (3) we get\n\\[\nabscissa\\,ordinate\\,(2abscissa\\,\\dot{abscissa} - 2ordinate\\,\\dot{ordinate}) + (abscissa\\,\\dot{ordinate} + \\dot{abscissa}\\,ordinate)\\left(1 - abscissa^{2} + ordinate^{2}\\right) = 0 .\n\\]\n\nThis is the differential equation of the suggested family of hyperbolas and it simplifies to (1) as required."
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "lampshade",
        "y": "sandstorm",
        "t": "waterfall",
        "A": "pendulum",
        "B": "quartzite"
      },
      "question": "8. A particle \\( (lampshade, sandstorm) \\) moves so that its angular velocities about \\( (1,0) \\) and \\( (-1,0) \\) are equal in magnitude but opposite in sign. Prove that\n\\[\nsandstorm\\left(lampshade^{2}+sandstorm^{2}+1\\right) d lampshade=lampshade\\left(lampshade^{2}+sandstorm^{2}-1\\right) d sandstorm\n\\]\nand verify that this is the differential equation of the family of rectangular hyperbolas passing through \\( (1,0) \\) and \\( (-1,0) \\) and having the origin as center.",
      "solution": "Solution. The angular velocity about the origin of a point moving according to the parametric equations\n\\[\nlampshade=lampshade(waterfall), \\quad sandstorm=sandstorm(waterfall)\n\\]\nis\n\\[\n\\frac{lampshade \\dot{sandstorm}-sandstorm \\dot{lampshade}}{lampshade^{2}+sandstorm^{2}}\n\\]\nwhere the dots indicate differentiation with respect to \\( waterfall \\). Translating the center of reference, first to \\( (1,0) \\) and then to \\( (-1,0) \\), we can express the condition as\n\\[\n\\frac{(lampshade-1) \\dot{sandstorm}-sandstorm \\dot{lampshade}}{(lampshade-1)^{2}+sandstorm^{2}}+\\frac{(lampshade+1) \\dot{sandstorm}-sandstorm \\dot{lampshade}}{(lampshade+1)^{2}+sandstorm^{2}}=0 .\n\\]\n\nFrom this we get\n\\[\n\\begin{array}{c}\n\\quad(lampshade \\dot{sandstorm}-sandstorm \\dot{lampshade}-\\dot{sandstorm})\\left(lampshade^{2}+sandstorm^{2}+1+2 lampshade\\right) \\\\\n+(lampshade \\dot{sandstorm}-sandstorm \\dot{lampshade}+\\dot{sandstorm})\\left(lampshade^{2}+sandstorm^{2}+1-2 lampshade\\right)=0,\n\\end{array}\n\\]\nwhich simplifies to\n\\[\nlampshade\\left(lampshade^{2}+sandstorm^{2}-1\\right) \\dot{sandstorm}=sandstorm\\left(lampshade^{2}+sandstorm^{2}+1\\right) \\dot{lampshade}\n\\]\n\nThis is equivalent to the required differential equation.\nA central rectangular hyperbola has an equation of the form\n\\[\npendulum\\left(lampshade^{2}-sandstorm^{2}\\right)+quartzite lampshade sandstorm=1\n\\]\n\nIt passes through \\( (-1,0) \\) and \\( (1,0) \\) if and only if \\( pendulum=1 \\). So the suggested family is given by\n\\[\nlampshade^{2}-sandstorm^{2}+quartzite lampshade sandstorm=1\n\\]\n\nDifferentiation with respect to \\( \\boldsymbol{waterfall} \\) yields\n\\[\n2 lampshade \\dot{lampshade}-2 sandstorm \\dot{sandstorm}+quartzite(lampshade \\dot{sandstorm}+\\dot{lampshade} sandstorm)=0 .\n\\]\n\nEliminating \\( quartzite \\) between (2) and (3) we get\n\\[\nlampshade sandstorm(2 lampshade \\dot{lampshade}-2 sandstorm \\dot{sandstorm})+(lampshade \\dot{sandstorm}+\\dot{lampshade} sandstorm)\\left(1-lampshade^{2}+sandstorm^{2}\\right)=0 .\n\\]\n\nThis is the differential equation of the suggested family of hyperbolas and it simplifies to (1) as required."
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "immobilevalue",
        "y": "steadyheight",
        "t": "timelesspar",
        "A": "inconstant",
        "B": "variable"
      },
      "question": "8. A particle \\( (immobilevalue, steadyheight) \\) moves so that its angular velocities about \\( (1,0) \\) and \\( (-1,0) \\) are equal in magnitude but opposite in sign. Prove that\n\\[\nsteadyheight\\left(immobilevalue^{2}+steadyheight^{2}+1\\right) d immobilevalue=immobilevalue\\left(immobilevalue^{2}+steadyheight^{2}-1\\right) d steadyheight\n\\]\nand verify that this is the differential equation of the family of rectangular hyperbolas passing through \\( (1,0) \\) and \\( (-1,0) \\) and having the origin as center.",
      "solution": "Solution. The angular velocity about the origin of a point moving according to the parametric equations\n\\[\nimmobilevalue=immobilevalue(timelesspar), \\quad steadyheight=steadyheight(timelesspar)\n\\]\nis\n\\[\n\\frac{immobilevalue \\dot{steadyheight}-steadyheight \\dot{immobilevalue}}{immobilevalue^{2}+steadyheight^{2}}\n\\]\nwhere the dots indicate differentiation with respect to \\( timelesspar \\). Translating the center of reference, first to \\( (1,0) \\) and then to \\( (-1,0) \\), we can express the condition as\n\\[\n\\frac{(immobilevalue-1) \\dot{steadyheight}-steadyheight \\dot{immobilevalue}}{(immobilevalue-1)^{2}+steadyheight^{2}}+\\frac{(immobilevalue+1) \\dot{steadyheight}-steadyheight \\dot{immobilevalue}}{(immobilevalue+1)^{2}+steadyheight^{2}}=0 .\n\\]\n\nFrom this we get\n\\[\n\\begin{array}{c}\n\\quad(immobilevalue \\dot{steadyheight}-steadyheight \\dot{immobilevalue}-\\dot{steadyheight})\\left(immobilevalue^{2}+steadyheight^{2}+1+2 immobilevalue\\right) \\\\\n+(immobilevalue \\dot{steadyheight}-steadyheight \\dot{immobilevalue}+\\dot{steadyheight})\\left(immobilevalue^{2}+steadyheight^{2}+1-2 immobilevalue\\right)=0,\n\\end{array}\n\\]\nwhich simplifies to\n\\[\nimmobilevalue\\left(immobilevalue^{2}+steadyheight^{2}-1\\right) \\dot{steadyheight}=steadyheight\\left(immobilevalue^{2}+steadyheight^{2}+1\\right) \\dot{immobilevalue}\n\\]\n\nThis is equivalent to the required differential equation.\nA central rectangular hyperbola has an equation of the form\n\\[\ninconstant\\left(immobilevalue^{2}-steadyheight^{2}\\right)+variable \\, immobilevalue \\, steadyheight=1\n\\]\n\nIt passes through \\( (-1,0) \\) and \\( (1,0) \\) if and only if \\( inconstant=1 \\). So the suggested family is given by\n\\[\nimmobilevalue^{2}-steadyheight^{2}+variable \\, immobilevalue \\, steadyheight=1\n\\]\n\nDifferentiation with respect to \\( \\boldsymbol{timelesspar} \\) yields\n\\[\n2 immobilevalue \\dot{immobilevalue}-2 steadyheight \\dot{steadyheight}+variable(immobilevalue \\dot{steadyheight}+\\dot{immobilevalue} steadyheight)=0 .\n\\]\n\nEliminating \\( variable \\) between (2) and (3) we get\n\\[\nimmobilevalue steadyheight(2 immobilevalue \\dot{immobilevalue}-2 steadyheight \\dot{steadyheight})+(immobilevalue \\dot{steadyheight}+\\dot{immobilevalue} steadyheight)\\left(1-immobilevalue^{2}+steadyheight^{2}\\right)=0 .\n\\]\n\nThis is the differential equation of the suggested family of hyperbolas and it simplifies to (1) as required."
    },
    "garbled_string": {
      "map": {
        "x": "qzxwvtnp",
        "y": "hjgrksla",
        "t": "mpkdjuri",
        "A": "cnvlyefo",
        "B": "kzphgwra"
      },
      "question": "8. A particle \\( (qzxwvtnp, hjgrksla) \\) moves so that its angular velocities about \\( (1,0) \\) and \\( (-1,0) \\) are equal in magnitude but opposite in sign. Prove that\n\\[\nhjgrksla\\left(qzxwvtnp^{2}+hjgrksla^{2}+1\\right) d qzxwvtnp = qzxwvtnp\\left(qzxwvtnp^{2}+hjgrksla^{2}-1\\right) d hjgrksla\n\\]\nand verify that this is the differential equation of the family of rectangular hyperbolas passing through \\( (1,0) \\) and \\( (-1,0) \\) and having the origin as center.",
      "solution": "Solution. The angular velocity about the origin of a point moving according to the parametric equations\n\\[\nqzxwvtnp=qzxwvtnp(mpkdjuri), \\quad hjgrksla=hjgrksla(mpkdjuri)\n\\]\nis\n\\[\n\\frac{qzxwvtnp \\dot{hjgrksla}-hjgrksla \\dot{qzxwvtnp}}{qzxwvtnp^{2}+hjgrksla^{2}}\n\\]\nwhere the dots indicate differentiation with respect to \\( mpkdjuri \\). Translating the center of reference, first to \\( (1,0) \\) and then to \\( (-1,0) \\), we can express the condition as\n\\[\n\\frac{(qzxwvtnp-1) \\dot{hjgrksla}-hjgrksla \\dot{qzxwvtnp}}{(qzxwvtnp-1)^{2}+hjgrksla^{2}}+\\frac{(qzxwvtnp+1) \\dot{hjgrksla}-hjgrksla \\dot{qzxwvtnp}}{(qzxwvtnp+1)^{2}+hjgrksla^{2}}=0 .\n\\]\n\nFrom this we get\n\\[\n\\begin{array}{c}\n\\quad(qzxwvtnp \\dot{hjgrksla}-hjgrksla \\dot{qzxwvtnp}-\\dot{hjgrksla})\\left(qzxwvtnp^{2}+hjgrksla^{2}+1+2 qzxwvtnp\\right) \\\\\n+(qzxwvtnp \\dot{hjgrksla}-hjgrksla \\dot{qzxwvtnp}+\\dot{hjgrksla})\\left(qzxwvtnp^{2}+hjgrksla^{2}+1-2 qzxwvtnp\\right)=0,\n\\end{array}\n\\]\nwhich simplifies to\n\\[\nqzxwvtnp\\left(qzxwvtnp^{2}+hjgrksla^{2}-1\\right) \\dot{hjgrksla}=hjgrksla\\left(qzxwvtnp^{2}+hjgrksla^{2}+1\\right) \\dot{qzxwvtnp}\n\\]\n\nThis is equivalent to the required differential equation.\nA central rectangular hyperbola has an equation of the form\n\\[\ncnvlyefo\\left(qzxwvtnp^{2}-hjgrksla^{2}\\right)+kzphgwra \\, qzxwvtnp \\, hjgrksla=1\n\\]\n\nIt passes through \\( (-1,0) \\) and \\( (1,0) \\) if and only if \\( cnvlyefo=1 \\). So the suggested family is given by\n\\[\nqzxwvtnp^{2}-hjgrksla^{2}+kzphgwra \\, qzxwvtnp \\, hjgrksla=1\n\\]\n\nDifferentiation with respect to \\( \\boldsymbol{mpkdjuri} \\) yields\n\\[\n2 qzxwvtnp \\dot{qzxwvtnp}-2 hjgrksla \\dot{hjgrksla}+kzphgwra(qzxwvtnp \\dot{hjgrksla}+\\dot{qzxwvtnp} hjgrksla)=0 .\n\\]\n\nEliminating \\( kzphgwra \\) between (2) and (3) we get\n\\[\nqzxwvtnp \\, hjgrksla(2 qzxwvtnp \\dot{qzxwvtnp}-2 hjgrksla \\dot{hjgrksla})+(qzxwvtnp \\dot{hjgrksla}+\\dot{qzxwvtnp} hjgrksla)\\left(1-qzxwvtnp^{2}+hjgrksla^{2}\\right)=0 .\n\\]\n\nThis is the differential equation of the suggested family of hyperbolas and it simplifies to (1) as required."
    },
    "kernel_variant": {
      "question": "Let a point–mass move in the plane and denote its position at time t by (x(t),y(t)).  Throughout the motion we assume \n\n(1)  the instantaneous angular velocities about the two fixed points (2,0) and (−2,0) are non–zero and have equal magnitude but opposite sign;\n\n(2)  the orbit never meets the y–axis, i.e.  x(t)≠0 for every t.\n\n(a)  Show that the coordinates of the particle satisfy the first–order differential equation\n        y\\bigl(x^{2}+y^{2}+4\\bigr)\\,dx = x\\bigl(x^{2}+y^{2}-4\\bigr)\\,dy.                 (∗)\n\n(b)  Prove that every trajectory that fulfils (1) and (2) is an entire branch of exactly one curve of the one–parameter family of rectangular hyperbolas\n        x^{2}-y^{2}+Bxy = 4 \\qquad (B \\text{ constant}),\nall of which are centred at the origin and pass through the points (2,0) and (−2,0).\n\nRemark.  If assumption (2) is dropped, the y–axis x=0 is an additional (singular) integral curve of (∗); if in (1) the phrase “non–zero” is omitted, the x–axis y=0 becomes another singular integral curve.",
      "solution": "Part (a)\nLet a superposed dot denote differentiation with respect to t.  The angular velocity of the moving point about an arbitrary point (a,0) is\n        ω_a = \\dfrac{(x-a)\\dot y - y\\dot x}{(x-a)^2+y^2}.\nBecause the angular velocities about (2,0) and (−2,0) are supposed to have the same magnitude but opposite sign,\n        \\frac{(x-2)\\dot y - y\\dot x}{(x-2)^2+y^2} + \\frac{(x+2)\\dot y - y\\dot x}{(x+2)^2+y^2}=0.         (1)\nIntroduce\n        S := x^2+y^2+4,\\qquad W := x\\dot y - y\\dot x.\nThen (x\\mp2)^2+y^2 = S\\mp4x and (x\\mp2)\\dot y - y\\dot x = W\\mp2\\dot y.  Multiplying (1) by (S-4x)(S+4x) and simplifying gives\n        2SW - 16 x\\dot y =0 \\;\\Longrightarrow\\; W S = 8x\\dot y.                       (2)\nFinally, separate the \\dot x– and \\dot y–terms:\n        x(x^2+y^2-4)\\dot y = y(x^2+y^2+4)\\dot x.\nReplacing \\dot x\\,dt by dx and \\dot y\\,dt by dy yields the differential equation (∗).\n\nPart (b)  –  the integral curves of (∗)\n1.  A whole one–parameter family of rectangular hyperbolas satisfies (∗).\n    \n    A central rectangular hyperbola can be written in the form\n           A(x^2-y^2)+Bxy=C,\\qquad C\\neq0.\n    Passing through (±2,0) forces C=4A.  Normalising with C=4 gives A=1 and the family\n           F(x,y;B):=x^2-y^2+ Bxy-4 =0.                                    (3)\n    Differentiating (3) with respect to t and eliminating B between (3) and its time–derivative one recovers exactly the velocity form of (∗); hence every member of (3) is an integral curve of (∗).\n\n2.  Conversely, let (x(t),y(t)) be an arbitrary solution of (∗) with x(t)≠0 (assumption (2)).  Set\n           H(x,y):=x^2-y^2-4,\\qquad G(x,y):=xy.\n    On every open sub-interval on which y(t)≠0 we define\n           B(t):=\\frac{H(x(t),y(t))}{G(x(t),y(t))}.                       (4)\n    (The quotient is well defined because both factors are non–zero there.)  We show that B(t) is constant and afterwards extend this constancy across any points where y=0 by continuity.\n\n    Differentiate (4) with respect to t; using \\dot H=2x\\dot x-2y\\dot y and \\dot G = x\\dot y+y\\dot x we obtain\n        G^{2}\\dot B = G(2x\\dot x-2y\\dot y) - H(x\\dot y + y\\dot x).          (5)\n    Denote S=x^{2}+y^{2}.  The velocity form of (∗) is\n        y(S+4)\\dot x = x(S-4)\\dot y.                                      (6)\n\n    A straightforward expansion of (5) gives\n        G^{2}\\dot B = y(S+4)\\dot x - x(S-4)\\dot y.                         (7)\n    Substituting (6) into (7) makes the right–hand side vanish identically, so\n        G^{2}\\dot B = 0.\n    On the intervals considered we have G=xy ≠0, hence \\dot B=0 there.  Thus B(t)=B_{0} is constant on each connected component of {t | y(t)≠0}.  Since both H and G are continuous functions, the ratio H/G extends continuously to points where y=0; therefore the same constant value B_{0} applies at those points as well.  Consequently the identity\n           x^{2}-y^{2}+B_{0}xy-4\\equiv0                                  (8)\n    holds for all t, i.e. the whole orbit lies on the fixed hyperbola F(x,y;B_{0})=0.  Because the motion is continuous, the orbit is an entire branch of that hyperbola.\n\nCombining the two directions, the integral curves of the Pfaffian form (∗) that satisfy conditions (1) and (2) are exactly the rectangular hyperbolas (3).\n\nSingular solutions.  In deriving (∗) we divided by x and by x\\dot y+y\\dot x, hence the lines x=0 and y=0 may appear as singular integral curves.\n•  If x=0 the angular velocities required in (1) are still non-zero, so the y-axis is an integral curve of (∗).  It is excluded by assumption (2).\n•  If y=0 the angular velocities about (±2,0) vanish; this possibility is excluded by the phrase “non-zero” in (1).\n\nTherefore, under the hypotheses imposed in the problem, the only admissible trajectories are the rectangular hyperbolas (3), as required.",
      "_meta": {
        "core_steps": [
          "Express the angular velocity about a point as (x ẏ – y ẋ)/(distance²).",
          "Impose that the angular velocities about the two symmetric points are equal in magnitude and opposite in sign, so their sum is zero.",
          "Algebraically clear denominators and collect terms to obtain the first-order differential equation y(x²+y²+1) dx = x(x²+y²−1) dy.",
          "Describe every rectangular hyperbola centered at the origin by A(x²−y²)+Bxy = 1 and use the requirement that it pass through the two fixed points to set A = 1.",
          "Differentiate that curve with respect to t, eliminate the parameter B, and confirm the resulting relation coincides with the previously derived differential equation."
        ],
        "mutable_slots": {
          "slot1": {
            "description": "Half-distance of the two fixed points from the origin (currently the ±1 in (1,0) and (−1,0); becomes ±a everywhere).",
            "original": "1"
          },
          "slot2": {
            "description": "Overall scaling on the hyperbola family (right-hand side of A(x²−y²)+Bxy = 1; any non-zero constant would work since it can be absorbed into A and B).",
            "original": "1"
          }
        }
      }
    }
  },
  "checked": true,
  "problem_type": "proof",
  "iteratively_fixed": true
}