{
  "index": "1942-B-2",
  "type": "ANA",
  "tag": [
    "ANA",
    "ALG"
  ],
  "difficulty": "",
  "question": "8. For the family of parabolas\n\\[\ny=\\frac{a^{3} x^{2}}{3}+\\frac{a^{2} x}{2}-2 a,\n\\]\n(i) find the locus of vertices,\n(ii) find the envelope,\n(iii) sketch the envelope and two typical curves of the family.",
  "solution": "Solution. (i) The given equation can be written in standard form as\n\\[\ny+\\frac{35}{16} a=\\frac{a^{3}}{3}\\left(x+\\frac{3}{4 a}\\right)^{2}\n\\]\nwhence a typical vertex is\n\\[\n\\left(-\\frac{3}{4 a},-\\frac{35}{16} a\\right)\n\\]\n\nIf \\( a=0 \\), then the given curve is a straight line, not a parabola, and therefore it has no vertex. The vertices of the parabolas in the system all lie on the hyperbola \\( x y=105 / 64 \\).\n\nConversely every point of this hyperbola is the vertex of a unique member of this family, since if \\( \\left(x_{0}, y_{0}\\right) \\) is on the hyperbola, then \\( x_{0}=-3 / 4 a \\), \\( y_{0}=-35 a / 16 \\) can be solved uniquely for \\( a \\).\n(ii) Let\n\\[\nf(x, y, a)=\\frac{a^{3} x^{2}}{3}+\\frac{a^{2} x}{2}-2 a-y\n\\]\n\nThen\n\\[\n\\frac{\\partial f}{\\partial a}(x, y, a)=(a x+2)(a x-1)\n\\]\n\nTo find the envelope of the family, we eliminate \\( a \\) between \\( \\partial f / \\partial a=0 \\) and \\( f=0 \\). This gives\n\\[\nx y=\\frac{(a x)^{3}}{3}+\\frac{(a x)^{2}}{2}-2 a x=\\frac{1}{3}+\\frac{1}{2}-2=\\frac{-7}{6}\n\\]\nor\n\\[\n=\\frac{-8}{3}+\\frac{4}{2}+4=\\frac{10}{3}\n\\]\ndepending on which of the two factors of \\( \\partial f / \\partial a \\) we choose to equate to zero.\n\nWe can readily verify that the parabola corresponding to the parameter \\( a \\) is tangent to the hyperbola\n\\[\nx y=-7 / 6 \\text { at }\\left(\\frac{1}{a},-\\frac{7 a}{6}\\right)\n\\]\nand tangent to the hyperbola\n\\[\nx y=10 / 3 \\text { at }\\left(-\\frac{2}{a},-\\frac{5}{3} a\\right) .\n\\]\n\nHence the envelope is the union of the two hyperbolas.",
  "vars": [
    "x",
    "y"
  ],
  "params": [
    "a",
    "x_0",
    "y_0"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "abscis",
        "y": "ordinate",
        "a": "paracon",
        "x_0": "abscisnaught",
        "y_0": "ordinatenaught"
      },
      "question": "8. For the family of parabolas\n\\[\nordinate=\\frac{paracon^{3} abscis^{2}}{3}+\\frac{paracon^{2} abscis}{2}-2 paracon,\n\\]\n(i) find the locus of vertices,\n(ii) find the envelope,\n(iii) sketch the envelope and two typical curves of the family.",
      "solution": "Solution. (i) The given equation can be written in standard form as\n\\[\nordinate+\\frac{35}{16} paracon=\\frac{paracon^{3}}{3}\\left(abscis+\\frac{3}{4 paracon}\\right)^{2}\n\\]\nwhence a typical vertex is\n\\[\n\\left(-\\frac{3}{4 paracon},-\\frac{35}{16} paracon\\right)\n\\]\n\nIf \\( paracon=0 \\), then the given curve is a straight line, not a parabola, and therefore it has no vertex. The vertices of the parabolas in the system all lie on the hyperbola \\( abscis \\, ordinate=105 / 64 \\).\n\nConversely every point of this hyperbola is the vertex of a unique member of this family, since if \\( \\left(abscisnaught, ordinatenaught\\right) \\) is on the hyperbola, then \\( abscisnaught=-3 / 4 paracon \\), \\( ordinatenaught=-35 paracon / 16 \\) can be solved uniquely for \\( paracon \\).\n(ii) Let\n\\[\nf(abscis, ordinate, paracon)=\\frac{paracon^{3} \\, abscis^{2}}{3}+\\frac{paracon^{2} \\, abscis}{2}-2 paracon-ordinate\n\\]\n\nThen\n\\[\n\\frac{\\partial f}{\\partial paracon}(abscis, ordinate, paracon)=(paracon \\, abscis+2)(paracon \\, abscis-1)\n\\]\n\nTo find the envelope of the family, we eliminate \\( paracon \\) between \\( \\partial f / \\partial paracon=0 \\) and \\( f=0 \\). This gives\n\\[\nabscis \\, ordinate=\\frac{(paracon \\, abscis)^{3}}{3}+\\frac{(paracon \\, abscis)^{2}}{2}-2 paracon \\, abscis=\\frac{1}{3}+\\frac{1}{2}-2=\\frac{-7}{6}\n\\]\nor\n\\[\n=\\frac{-8}{3}+\\frac{4}{2}+4=\\frac{10}{3}\n\\]\ndepending on which of the two factors of \\( \\partial f / \\partial paracon \\) we choose to equate to zero.\n\nWe can readily verify that the parabola corresponding to the parameter \\( paracon \\) is tangent to the hyperbola\n\\[\nabscis \\, ordinate=-7 / 6 \\text { at }\\left(\\frac{1}{paracon},-\\frac{7 paracon}{6}\\right)\n\\]\nand tangent to the hyperbola\n\\[\nabscis \\, ordinate=10 / 3 \\text { at }\\left(-\\frac{2}{paracon},-\\frac{5}{3} paracon\\right) .\n\\]\n\nHence the envelope is the union of the two hyperbolas."
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "sunflower",
        "y": "pineapple",
        "a": "marathon",
        "x_0": "rainstorm",
        "y_0": "butterfly"
      },
      "question": "8. For the family of parabolas\n\\[\npineapple=\\frac{marathon^{3} sunflower^{2}}{3}+\\frac{marathon^{2} sunflower}{2}-2 marathon,\n\\]\n(i) find the locus of vertices,\n(ii) find the envelope,\n(iii) sketch the envelope and two typical curves of the family.",
      "solution": "Solution. (i) The given equation can be written in standard form as\n\\[\npineapple+\\frac{35}{16} marathon=\\frac{marathon^{3}}{3}\\left(sunflower+\\frac{3}{4 marathon}\\right)^{2}\n\\]\nwhence a typical vertex is\n\\[\n\\left(-\\frac{3}{4 marathon},-\\frac{35}{16} marathon\\right)\n\\]\n\nIf \\( marathon=0 \\), then the given curve is a straight line, not a parabola, and therefore it has no vertex. The vertices of the parabolas in the system all lie on the hyperbola \\( sunflower pineapple=105 / 64 \\).\n\nConversely every point of this hyperbola is the vertex of a unique member of this family, since if \\( \\left(rainstorm, butterfly\\right) \\) is on the hyperbola, then \\( rainstorm=-3 / 4 marathon \\), \\( butterfly=-35 marathon / 16 \\) can be solved uniquely for \\( marathon \\).\n\n(ii) Let\n\\[\nf(sunflower, pineapple, marathon)=\\frac{marathon^{3} sunflower^{2}}{3}+\\frac{marathon^{2} sunflower}{2}-2 marathon-pineapple\n\\]\n\nThen\n\\[\n\\frac{\\partial f}{\\partial marathon}(sunflower, pineapple, marathon)=(marathon sunflower+2)(marathon sunflower-1)\n\\]\n\nTo find the envelope of the family, we eliminate \\( marathon \\) between \\( \\partial f / \\partial marathon=0 \\) and \\( f=0 \\). This gives\n\\[\nsunflower pineapple=\\frac{(marathon sunflower)^{3}}{3}+\\frac{(marathon sunflower)^{2}}{2}-2 marathon sunflower=\\frac{1}{3}+\\frac{1}{2}-2=\\frac{-7}{6}\n\\]\nor\n\\[\n=\\frac{-8}{3}+\\frac{4}{2}+4=\\frac{10}{3}\n\\]\ndepending on which of the two factors of \\( \\partial f / \\partial marathon \\) we choose to equate to zero.\n\nWe can readily verify that the parabola corresponding to the parameter \\( marathon \\) is tangent to the hyperbola\n\\[\nsunflower pineapple=-7 / 6 \\text { at }\\left(\\frac{1}{marathon},-\\frac{7 marathon}{6}\\right)\n\\]\nand tangent to the hyperbola\n\\[\nsunflower pineapple=10 / 3 \\text { at }\\left(-\\frac{2}{marathon},-\\frac{5}{3} marathon\\right) .\n\\]\n\nHence the envelope is the union of the two hyperbolas."
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "verticalaxis",
        "y": "horizontalaxis",
        "a": "resultvalue",
        "x_0": "slopevalue",
        "y_0": "interceptvalue"
      },
      "question": "8. For the family of parabolas\n\\[\nhorizontalaxis=\\frac{resultvalue^{3} verticalaxis^{2}}{3}+\\frac{resultvalue^{2} verticalaxis}{2}-2 resultvalue,\n\\]\n(i) find the locus of vertices,\n(ii) find the envelope,\n(iii) sketch the envelope and two typical curves of the family.",
      "solution": "Solution. (i) The given equation can be written in standard form as\n\\[\nhorizontalaxis+\\frac{35}{16} resultvalue=\\frac{resultvalue^{3}}{3}\\left(verticalaxis+\\frac{3}{4 resultvalue}\\right)^{2}\n\\]\nwhence a typical vertex is\n\\[\n\\left(-\\frac{3}{4 resultvalue},-\\frac{35}{16} resultvalue\\right)\n\\]\n\nIf \\( resultvalue=0 \\), then the given curve is a straight line, not a parabola, and therefore it has no vertex. The vertices of the parabolas in the system all lie on the hyperbola \\( verticalaxis horizontalaxis=105 / 64 \\).\n\nConversely every point of this hyperbola is the vertex of a unique member of this family, since if \\( \\left(slopevalue, interceptvalue\\right) \\) is on the hyperbola, then \\( slopevalue=-3 / 4 resultvalue \\), \\( interceptvalue=-35 resultvalue / 16 \\) can be solved uniquely for \\( resultvalue \\).\n(ii) Let\n\\[\nf(verticalaxis, horizontalaxis, resultvalue)=\\frac{resultvalue^{3} verticalaxis^{2}}{3}+\\frac{resultvalue^{2} verticalaxis}{2}-2 resultvalue-horizontalaxis\n\\]\n\nThen\n\\[\n\\frac{\\partial f}{\\partial resultvalue}(verticalaxis, horizontalaxis, resultvalue)=(resultvalue verticalaxis+2)(resultvalue verticalaxis-1)\n\\]\n\nTo find the envelope of the family, we eliminate \\( resultvalue \\) between \\( \\partial f / \\partial resultvalue=0 \\) and \\( f=0 \\). This gives\n\\[\nverticalaxis horizontalaxis=\\frac{(resultvalue verticalaxis)^{3}}{3}+\\frac{(resultvalue verticalaxis)^{2}}{2}-2 resultvalue verticalaxis=\\frac{1}{3}+\\frac{1}{2}-2=\\frac{-7}{6}\n\\]\nor\n\\[\n=\\frac{-8}{3}+\\frac{4}{2}+4=\\frac{10}{3}\n\\]\ndepending on which of the two factors of \\( \\partial f / \\partial resultvalue \\) we choose to equate to zero.\n\nWe can readily verify that the parabola corresponding to the parameter \\( resultvalue \\) is tangent to the hyperbola\n\\[\nverticalaxis horizontalaxis=-7 / 6 \\text { at }\\left(\\frac{1}{resultvalue},-\\frac{7 resultvalue}{6}\\right)\n\\]\nand tangent to the hyperbola\n\\[\nverticalaxis horizontalaxis=10 / 3 \\text { at }\\left(-\\frac{2}{resultvalue},-\\frac{5}{3} resultvalue\\right) .\n\\]\n\nHence the envelope is the union of the two hyperbolas."
    },
    "garbled_string": {
      "map": {
        "x": "qzxwvtnp",
        "y": "hjgrksla",
        "a": "mcfpdqer",
        "x_0": "lkjhgfds",
        "y_0": "poiuytre"
      },
      "question": "8. For the family of parabolas\n\\[\nhjgrksla=\\frac{mcfpdqer^{3} qzxwvtnp^{2}}{3}+\\frac{mcfpdqer^{2} qzxwvtnp}{2}-2 mcfpdqer,\n\\]\n(i) find the locus of vertices,\n(ii) find the envelope,\n(iii) sketch the envelope and two typical curves of the family.",
      "solution": "Solution. (i) The given equation can be written in standard form as\n\\[\nhjgrksla+\\frac{35}{16} mcfpdqer=\\frac{mcfpdqer^{3}}{3}\\left(qzxwvtnp+\\frac{3}{4 mcfpdqer}\\right)^{2}\n\\]\nwhence a typical vertex is\n\\[\n\\left(-\\frac{3}{4 mcfpdqer},-\\frac{35}{16} mcfpdqer\\right)\n\\]\n\nIf \\( mcfpdqer=0 \\), then the given curve is a straight line, not a parabola, and therefore it has no vertex. The vertices of the parabolas in the system all lie on the hyperbola \\( qzxwvtnp hjgrksla=105 / 64 \\).\n\nConversely every point of this hyperbola is the vertex of a unique member of this family, since if \\( \\left(lkjhgfds, poiuytre\\right) \\) is on the hyperbola, then \\( lkjhgfds=-3 / 4 mcfpdqer \\), \\( poiuytre=-35 mcfpdqer / 16 \\) can be solved uniquely for \\( mcfpdqer \\).\n(ii) Let\n\\[\nf(qzxwvtnp, hjgrksla, mcfpdqer)=\\frac{mcfpdqer^{3} qzxwvtnp^{2}}{3}+\\frac{mcfpdqer^{2} qzxwvtnp}{2}-2 mcfpdqer-hjgrksla\n\\]\n\nThen\n\\[\n\\frac{\\partial f}{\\partial mcfpdqer}(qzxwvtnp, hjgrksla, mcfpdqer)=(mcfpdqer qzxwvtnp+2)(mcfpdqer qzxwvtnp-1)\n\\]\n\nTo find the envelope of the family, we eliminate \\( mcfpdqer \\) between \\( \\partial f / \\partial mcfpdqer=0 \\) and \\( f=0 \\). This gives\n\\[\nqzxwvtnp hjgrksla=\\frac{(mcfpdqer qzxwvtnp)^{3}}{3}+\\frac{(mcfpdqer qzxwvtnp)^{2}}{2}-2 mcfpdqer qzxwvtnp=\\frac{1}{3}+\\frac{1}{2}-2=\\frac{-7}{6}\n\\]\nor\n\\[\n=\\frac{-8}{3}+\\frac{4}{2}+4=\\frac{10}{3}\n\\]\ndepending on which of the two factors of \\( \\partial f / \\partial mcfpdqer \\) we choose to equate to zero.\n\nWe can readily verify that the parabola corresponding to the parameter \\( mcfpdqer \\) is tangent to the hyperbola\n\\[\nqzxwvtnp hjgrksla=-7 / 6 \\text { at }\\left(\\frac{1}{mcfpdqer},-\\frac{7 mcfpdqer}{6}\\right)\n\\]\nand tangent to the hyperbola\n\\[\nqzxwvtnp hjgrksla=10 / 3 \\text { at }\\left(-\\frac{2}{mcfpdqer},-\\frac{5}{3} mcfpdqer\\right) .\n\\]\n\nHence the envelope is the union of the two hyperbolas."
    },
    "kernel_variant": {
      "question": "Let  \n\\[\nS_{a}\\;:\\;\nz=\\frac{a^{3}}{3}\\bigl(x^{2}+y^{2}\\bigr)\n+\\frac{a^{2}}{2}\\,(x-2y)-6a ,\n\\qquad a\\in\\mathbb R\\setminus\\{0\\},\n\\tag{$\\star$}\n\\]\nbe a one–parameter family of quadratic surfaces in $\\mathbb R^{3}$.  \nThroughout write  \n\\[\nr^{2}=x^{2}+y^{2},\n\\qquad\ns=x-2y .\n\\]\n\n(i)  A \\emph{vertex} of $S_{a}$ is a point for which the gradient (with respect to $x,y$) of the right–hand side of $(\\star)$ vanishes.  \nDetermine all vertices and describe their locus $\\mathcal C$ both parametrically and in Cartesian form.\n\n(ii)  Denote by $\\Sigma$ the \\emph{envelope} of the family\n$(\\star)$, i.e. the set of points that are tangent to (at least) one\nmember $S_{a}$.\n\n\\quad (a)  Show that $(x,y,z)\\in\\Sigma$ iff the system  \n\\[\nF(x,y,z,a)=0,\\qquad F_{a}(x,y,z,a)=0,\\qquad\nF:=z-\\frac{a^{3}}{3}r^{2}-\\frac{a^{2}}{2}s+6a\n\\tag{E}\n\\]\nadmits a real solution $a$.\n\n\\quad (b)  Solving $F_{a}=0$ for $r^{2}$ and using the polar\nrepresentation $(x,y)=(r\\cos\\theta,r\\sin\\theta)$,\nobtain a smooth parametrisation\n\\[\nX(a,\\theta)=\\bigl(x(a,\\theta),y(a,\\theta),z(a,\\theta)\\bigr),\n\\qquad a\\in\\mathbb R\\setminus\\{0\\},\\; \\theta\\in[0,2\\pi),\n\\]\nof $\\Sigma$.\n\n\\quad (c)  Eliminate $a$ from the two equations in $(E)$ and show that\n$\\Sigma$ is the zero–locus of the irreducible sextic\n\\[\nP(x,y,z)=\n6z^{2}(x^{2}+y^{2})^{2}-z(x-2y)^{3}\n-36z(x^{2}+y^{2})(x-2y)\n-18(x-2y)^{2}-576(x^{2}+y^{2}).\n\\tag{$\\dagger$}\n\\]\n\n(iii)  Fix $a\\neq 0$ and set\n\\[\n\\Gamma_{a}:=S_{a}\\cap\\Sigma .\n\\]\n\n\\quad (a)  Show that $\\Gamma_{a}$ is a smooth closed space curve.\n\n\\quad (b)  Prove that the orthogonal projection of\n$\\Gamma_{a}$ onto the $(x,y)$–plane is the circle\n\\[\n\\left(x+\\frac{1}{2a}\\right)^{2}+\n\\left(y-\\frac{1}{a}\\right)^{2}=\\frac{29}{4a^{2}},\n\\tag{$\\diamondsuit$}\n\\]\nand produce an explicit parametrisation of $\\Gamma_{a}$ valid for all\n$a\\neq 0$.\n\n\\quad (c)  Combine the two defining equations of $\\Gamma_{a}$ and deduce\nthat every point of $\\Gamma_{a}$ satisfies the additional \\emph{linear}\nrelation\n\\[\n6z-a^{2}\\bigl(x-2y\\bigr)+24a=0.\n\\]\nHence each $\\Gamma_{a}$ is contained in the plane\n\\[\n\\Pi_{a}:\\quad 6z-a^{2}(x-2y)+24a=0,\n\\]\nand therefore planar.  Show further that $\\Gamma_{a}$ is not a line\n(e.g.\\ by using the result of (b)).\n\n--------------------------------------------------------------------",
      "solution": "\\textbf{(i)  Vertices.}  \n\\[\n\\frac{\\partial z}{\\partial x}\n=\\frac{2a^{3}}{3}x+\\frac{a^{2}}{2},\n\\qquad\n\\frac{\\partial z}{\\partial y}\n=\\frac{2a^{3}}{3}y-a^{2}.\n\\]\nBoth vanish exactly when \n\\[\nx=-\\frac{3}{4a},\\qquad y=\\frac{3}{2a}.\n\\]\nSubstituting in $(\\star)$ gives  \n\\[\nz=\\frac{a^{3}}{3}\\Bigl(\\frac{9}{16a^{2}}+\\frac{9}{4a^{2}}\\Bigr)\n    +\\frac{a^{2}}{2}\\Bigl(-\\frac{3}{4a}-\\frac{3}{a}\\Bigr)-6a\n  =-\\frac{111}{16}\\,a .\n\\]\nHence  \n\\[\n\\boxed{\\;\n\\mathcal C:\\;\n(x,y,z)=\\Bigl(-\\dfrac{3}{4a},\\,\\dfrac{3}{2a},\\,-\\dfrac{111}{16}a\\Bigr),\n\\qquad a\\in\\mathbb R\\setminus\\{0\\}\\;}.\n\\]\nEliminating $a$ yields the Cartesian description  \n\\[\ny+2x=0,\\qquad 64xz=333,\n\\]\na rectangular hyperbola contained in the plane $y+2x=0$.\n\n\\bigskip\n\\textbf{(ii)  The envelope $\\Sigma$.}\n\n\\emph{(a)}  A point belongs to the envelope precisely when it lies on\nsome $S_{a}$ \\emph{and} that surface is tangent there; the two\nconditions are $F=0$ and $F_{a}=0$, i.e. system $(E)$.\n\n\\smallskip\n\\emph{(b)  Explicit parametrisation.}\nFrom $F_{a}=0$ one obtains\n\\[\na^{2}r^{2}+as-6=0.\n\\tag{1}\n\\]\nWrite $(x,y)=(r\\cos\\theta,r\\sin\\theta)$ and put  \n\\[\nk(\\theta)=\\cos\\theta-2\\sin\\theta.\n\\]\nEquation (1) is quadratic in $r$:\n\\[\na^{2}r^{2}+a r\\,k(\\theta)-6=0\n\\quad(r>0).\n\\]\nWith $\\varepsilon=\\operatorname{sgn}(a)$ its positive root is\n\\[\nr(a,\\theta)=\\frac{-\\varepsilon k(\\theta)+\\sqrt{k(\\theta)^{2}+24}}{2|a|}.\n\\tag{2}\n\\]\nUsing $s=r k(\\theta)$ and (2) we obtain the explicit map\n\\[\n\\boxed{\\,%\nX(a,\\theta)=\n\\Bigl(r\\cos\\theta,\\;\n      r\\sin\\theta,\\;\n      -4a+\\frac{a^{2}r\\,k(\\theta)}{6}\\Bigr)\\!,\n\\qquad\na\\neq 0,\\;0\\le\\theta<2\\pi\\,}.\n\\]\nAll three components depend smoothly on $(a,\\theta)$ and\n$\\partial_{\\theta}X,\\partial_{a}X$ are linearly independent, hence $X$\nis a $C^{\\infty}$ regular parametrisation of $\\Sigma$.\n\n\\smallskip\n\\emph{(c)  Implicit equation and irreducibility.}  \nSet\n\\[\nG_{1}=a^{2}r^{2}+as-6,\\qquad\nG_{2}=2a^{3}r^{2}+3a^{2}s-36a-6z .\n\\]\nEliminating $a$ from $G_{1}=G_{2}=0$ (e.g. by solving $a^{2}$ from\n$G_{1}$ and substituting into $G_{2}$) yields\n\\[\nP(x,y,z)=\n6z^{2}(x^{2}+y^{2})^{2}-z(x-2y)^{3}\n-36z(x^{2}+y^{2})(x-2y)\n-18(x-2y)^{2}-576(x^{2}+y^{2}) .\n\\]\nWe must still prove that $P$ is irreducible in\n$\\mathbb R[x,y,z]$.  \nPut $R=x^{2}+y^{2}$ and $S=x-2y$; then\n\\[\nP=6R^{2}z^{2}-\\bigl(S^{3}+36RS\\bigr)z-18S^{2}-576R.\n\\tag{3}\n\\]\n\\emph{Proof of irreducibility.}  \nEquation (3) is quadratic in $z$.  \nAssume that $P$ factors:\n\\[\nP=(u z+v)(w z+t),\\qquad u,v,w,t\\in\\mathbb R[x,y],\n\\]\nso that\n\\[\nu w = 6R^{2},\\qquad\nu t+v w = -(S^{3}+36RS),\\qquad\nv t = -18S^{2}-576R.\n\\tag{4}\n\\]\n\nBecause $R=x^{2}+y^{2}$ is irreducible in $\\mathbb R[x,y]$,\nthe first relation in (4) forces\n\\[\nu=c\\,R^{k},\\qquad w=d\\,R^{2-k},\n\\quad k\\in\\{0,1,2\\},\\; c d = 6,\n\\tag{5}\n\\]\nwith constants $c,d\\in\\mathbb R\\setminus\\{0\\}$.\n\n\\textbf{Case $k=0$.}  \nThen $u=c$ is constant and $w=d R^{2}$.  \nIn the second relation of (4) the term $v w$ is divisible by $R$,\nwhereas $u t=c\\,t$ is not (unless $t$ is divisible by $R$).\nSince the right–hand side $-(S^{3}+36RS)$ is \\emph{not} divisible\nby $R$ (because of the $S^{3}$ term), contradiction.\n\n\\textbf{Case $k=2$.}  \nSymmetric to $k=0$ (exchange $u$ and $w$) and leads to the same\ncontradiction.\n\n\\textbf{Case $k=1$.}  \nNow $u=cR$ and $w=dR$.  Then the sum $u t+v w$ in (4) is divisible by\n$R$, yet $-(S^{3}+36RS)$ is again not divisible by $R$.  Contradiction.\n\nAll possibilities are exhausted, hence $P$ cannot factor, and is\nirreducible in $\\mathbb R[x,y,z]$.\n\n\\bigskip\n\\textbf{(iii)  The curve $\\Gamma_{a}$.}\n\nWrite  \n\\[\nG_{1}(x,y)=a^{2}r^{2}+as-6,\\qquad\nG_{2}(x,y,z)=z-\\frac{a^{3}}{3}r^{2}-\\frac{a^{2}}{2}s+6a .\n\\]\n\n\\emph{(a)  Regularity and closedness.}  \nGradients of the two surfaces are  \n\\[\n\\nabla G_{1}=\\bigl(2a^{2}x+a,\\;2a^{2}y-2a,\\;0\\bigr),\n\\]\n\\[\n\\nabla G_{2}=\\Bigl(-\\tfrac{2a^{3}}{3}x-\\tfrac{a^{2}}{2},\\;\n                  -\\tfrac{2a^{3}}{3}y+a^{2},\\;1\\Bigr).\n\\]\nTheir cross product is\n\\[\n\\nabla G_{1}\\times\\nabla G_{2}\n   =\\bigl(2a^{2}y-2a,\\;\n          -\\!(2a^{2}x+a),\\;\n          (2a^{2}x+a)\\bigl(-\\tfrac{2a^{3}}{3}y+a^{2}\\bigr)\n          -(2a^{2}y-2a)\\bigl(-\\tfrac{2a^{3}}{3}x-\\tfrac{a^{2}}{2}\\bigr)\\bigr).\n\\]\nThe first two components vanish simultaneously only when\n\\[\nx=-\\frac{1}{2a},\\qquad y=\\frac{1}{a}.\n\\]\nAt those points\n\\(\nG_{1}=a^{2}\\bigl(\\tfrac{5}{4a^{2}}\\bigr)+a\\bigl(-\\tfrac{5}{2a}\\bigr)-6=-\\tfrac{29}{4}\\neq 0,\n\\)\nso they are \\emph{not} on $\\Gamma_{a}$.  \nConsequently $\\nabla G_{1}$ and $\\nabla G_{2}$ are linearly\nindependent everywhere on $\\Gamma_{a}$, proving that\n$\\Gamma_{a}$ is a smooth one–dimensional submanifold of $\\mathbb R^{3}$,\ni.e. a smooth space curve.  \nBecause the parametrisation constructed below is $2\\pi$–periodic,\n$\\Gamma_{a}$ is compact and hence closed.\n\n\\smallskip\n\\emph{(b)  Orthogonal projection and parametrisation.}  \nFrom $G_{1}=0$ we have\n\\(\na^{2}r^{2}+as=6.\n\\)\nCompleting the square gives\n\\[\n\\left(x+\\frac{1}{2a}\\right)^{2}+\n\\left(y-\\frac{1}{a}\\right)^{2}\n      =\\frac{29}{4a^{2}},\n\\]\ni.e. the projection of $\\Gamma_{a}$ is the circle $(\\diamondsuit)$.\nPut\n\\[\n\\rho=\\frac{\\sqrt{29}}{2|a|},\\qquad\n\\varepsilon=\\operatorname{sgn}(a).\n\\]\nA $2\\pi$–periodic parametrisation of the circle is\n\\[\nx(\\phi)=-\\frac{1}{2a}+\\rho\\cos\\phi,\\qquad\ny(\\phi)=\\frac{1}{a}+\\rho\\sin\\phi,\n\\qquad 0\\le\\phi<2\\pi .\n\\]\nSubstituting into $G_{2}=0$ yields\n\\[\nz(\\phi)=-\\frac{53}{12}a\n        +\\frac{|a|\\sqrt{29}}{12}\\bigl(\\cos\\phi-2\\sin\\phi\\bigr).\n\\]\nHence\n\\[\n\\boxed{%\n\\Gamma_{a}:\\;\n\\Phi_{a}(\\phi)=\n\\Bigl(\n-\\tfrac{1}{2a}+\\rho\\cos\\phi,\\;\n\\tfrac{1}{a}+\\rho\\sin\\phi,\\;\n-\\tfrac{53}{12}a+\n\\tfrac{|a|\\sqrt{29}}{12}\\,(\\cos\\phi-2\\sin\\phi)\n\\Bigr),\\quad 0\\le\\phi<2\\pi\\;}\n\\]\nparametrises the whole curve.\n\n\\smallskip\n\\emph{(c)  Planarity and non-linearity.}  \nTo eliminate the $r^{2}$ term, multiply $G_{1}=0$ by $\\dfrac{a}{3}$ and add the result to $G_{2}=0$:\n\\[\n\\frac{a}{3}G_{1}+G_{2}\n=z-\\frac{a^{2}}{6}s+4a=0 .\n\\]\nRearranging gives the promised linear relation\n\\[\n6z-a^{2}s+24a=0.\n\\]\nThus every point of $\\Gamma_{a}$ lies in the plane\n\\[\n\\Pi_{a}:\\quad 6z-a^{2}(x-2y)+24a=0.\n\\]\nSince the orthogonal projection of $\\Gamma_{a}$ is the \\emph{circle}\n$(\\diamondsuit)$, $\\Gamma_{a}$ cannot be a straight line\n(the projection of a line onto the $(x,y)$–plane is a line).\nTherefore $\\Gamma_{a}$ is planar but not rectilinear.\n\n\\bigskip",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.394978",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher dimension – the problem moves from planar curves to quadratic surfaces in ℝ³; vertices are now points in space, the envelope is a surface, and tangency occurs along curves instead of points.  \n\n2. Additional constraints – part (iii) forces the solver to locate and parametrize the two parabolic contact curves and to identify the orthogonal planes that contain them.  \n\n3. Deeper theory – computing the envelope requires solving a coupled system of a quartic and a cubic in the parameter 𝑎, finding a sextic resultant, and proving irreducibility.  Differential–geometric tools (first and second fundamental forms, Gaussian curvature, classification of parabolic points) are indispensable in part (iv).  \n\n4. Multiple interacting concepts – algebraic elimination, space curve geometry, surface envelopes, and differential geometry all appear and interact.  \n\n5. Length and subtlety – each sub–question demands several non-trivial steps; even locating the vertices requires working in two spatial directions, and the curvature computation obliges the solver to carry through a substantial symbolic calculation.\n\nAll these features make the enhanced kernel variant substantially more sophisticated and arduous than both the original textbook question and the current kernel variant confined to planar parabolas."
      }
    },
    "original_kernel_variant": {
      "question": "Let  \n\\[\nS_{a}\\;:\\;\nz=\\frac{a^{3}}{3}\\bigl(x^{2}+y^{2}\\bigr)\n+\\frac{a^{2}}{2}\\,(x-2y)-6a,\n\\qquad a\\in\\mathbb R\\setminus\\{0\\},\n\\tag{$\\star$}\n\\]  \nbe a one–parameter family of quadratic surfaces in $\\mathbb R^{3}$.  \nThroughout write  \n\\[\nr^{2}=x^{2}+y^{2}, \\qquad s=x-2y .\n\\]\n\n(i)  A vertex of $S_{a}$ is a point at which the gradient of the right–hand\nside of $(\\star)$ with respect to $(x,y)$ vanishes.  \nFind all vertices and determine their locus $\\mathcal C$ in $\\mathbb R^{3}$,\ngiving both a parametric and a Cartesian description.\n\n(ii)  Denote by $\\Sigma$ the envelope of the family $(\\star)$, i.e. the set\nof points that are tangent to at least one member $S_{a}$.\n\n\\quad (a)  Show that a point $(x,y,z)$ is on $\\Sigma$ if and only if the system  \n\\[\nF(x,y,z,a)=0,\\qquad F_{a}(x,y,z,a)=0,\\qquad  \nF:=z-\\frac{a^{3}}{3}r^{2}-\\frac{a^{2}}{2}s+6a\n\\tag{E}\n\\]\nadmits a real solution $a$.\n\n\\quad (b)  By solving $F_{a}=0$ for $r$ and using the polar\ndecomposition $(x,y)=(r\\cos\\theta,r\\sin\\theta)$, obtain a smooth  \nparametrisation  \n\\[\nX(a,\\theta)=\\bigl(x(a,\\theta),y(a,\\theta),z(a,\\theta)\\bigr),\n\\qquad a\\in\\mathbb R\\setminus\\{0\\},\\; \\theta\\in\\bigl[0,2\\pi\\bigr),\n\\]\nof $\\Sigma$.\n\n\\quad (c)  Eliminate $a$ from the two equations in (E) and prove that\n$\\Sigma$ is the zero–locus of the irreducible sextic\n\\[\nP(x,y,z)=\n6z^{2}(x^{2}+y^{2})^{2}-z(x-2y)^{3}\n-36z(x^{2}+y^{2})(x-2y)\n-18(x-2y)^{2}-576(x^{2}+y^{2})=0.\n\\tag{$\\dagger$}\n\\]\n\n(iii)  Fix $a\\neq 0$ and set\n\\[\n\\Gamma_{a}=S_{a}\\cap\\Sigma .\n\\]\n\n\\quad (a)  Show that $\\Gamma_{a}$ is a smooth closed space curve.  \n\n\\quad (b)  Prove that the orthogonal projection of\n$\\Gamma_{a}$ onto the $(x,y)$–plane is the circle\n\\[\n\\left(x+\\frac{1}{2a}\\right)^{2}+\n\\left(y-\\frac{1}{a}\\right)^{2}=\\frac{29}{4a^{2}}\\!,\n\\tag{$\\diamondsuit$}\n\\]\nand deduce an explicit parametrisation of $\\Gamma_{a}$ that is valid for\nall $a\\neq 0$.\n\n\\quad (c)  Show that $\\Gamma_{a}$ is not contained in any plane.\n\n(iv)  Using the parametrisation obtained in (ii)(b) compute the Gaussian\ncurvature $K$ of $\\Sigma$.\nShow that\n\\[\nK(x,y,z)=\n-\\frac{s^{2}}{4\\bigl(r^{2}+s^{2}\\bigr)^{3}},\n\\qquad s=x-2y,\n\\tag{$\\heartsuit$}\n\\]\nand conclude that the set\n\\[\n\\Gamma=\\Sigma\\cap\\{(x,y,z)\\mid x-2y=0\\}\n\\]\nconsists precisely of parabolic points of the envelope.\n\n────────────────────────────────────────────",
      "solution": "(i)  Vertices.  \n\\[\n\\frac{\\partial z}{\\partial x}=\\frac{2a^{3}}{3}x+\\frac{a^{2}}{2},\n\\qquad\n\\frac{\\partial z}{\\partial y}=\\frac{2a^{3}}{3}y-a^{2}.\n\\]\nSolving $\\partial z/\\partial x=\\partial z/\\partial y=0$ gives  \n\\[\nx=-\\frac{3}{4a},\\qquad y=\\frac{3}{2a}.\n\\]\nSubstitution in $(\\star)$ yields  \n\\[\nz=-\\frac{111}{16}\\,a.\n\\]\nHence the vertex curve is  \n\\[\n\\boxed{\\;\n\\mathcal C:\\;\n(x(a),y(a),z(a))=\n\\Bigl(-\\frac{3}{4a},\\frac{3}{2a},-\\frac{111}{16}a\\Bigr),\\\na\\in\\mathbb R\\setminus\\{0\\}\\;}\n\\]\nand eliminating $a$ gives  \n\\[\ny+2x=0,\\qquad 64xz=333,\n\\]\ni.e. $\\mathcal C$ is a rectangular hyperbola contained in the plane $y+2x=0$.\n\n(ii)  The envelope $\\Sigma$.\n\n(a)  A point belongs to the envelope iff it lies on some\nmember $S_{a}$ ($F=0$) and that member is tangent to the point\n($F_{a}=0$).  This is precisely the system (E).\n\n(b)  From $F_{a}=0$ we obtain  \n\\[\na^{2}r^{2}+as-6=0\\quad\\Longrightarrow\\quad\nr^{2}= \\frac{6-as}{a^{2}}, \\; r\\ge 0.\n\\]\nWith $(x,y)=(r\\cos\\theta,r\\sin\\theta)$,\n\\[\nr(a,\\theta)=\\frac{-a(\\cos\\theta-2\\sin\\theta)+\n\\sqrt{a^{2}(\\cos\\theta-2\\sin\\theta)^{2}+24a^{2}}}{2a^{2}},\n\\]\nthe positive root of the quadratic.  \nUsing $F=0$ one finds\n\\[\nz(a,\\theta)=\n-\\frac{a^{3}}{6}\\,r(a,\\theta)^{2}-3a.\n\\]\nThus\n\\[\n\\boxed{\\;\nX(a,\\theta)=\n\\bigl(r\\cos\\theta,\\;r\\sin\\theta,\\;\n      -\\tfrac{a^{3}}{6}r^{2}-3a\\bigr)\n,\\;\na\\neq 0,\\ \\theta\\in[0,2\\pi)\\;}\n\\]\nis a $C^{\\infty}$–parametrisation of $\\Sigma$.\n\n(c)  Set  \n\\[\nG_{1}:=a^{2}r^{2}+as-6,\\qquad\nG_{2}:=2a^{3}r^{2}+3a^{2}s-36a-6z .\n\\]\nEliminating $a$ from $\\{G_{1}=0,G_{2}=0\\}$ (e.g. by\nsubstituting $a^{2}$ and $a^{3}=a\\cdot a^{2}$ from $G_{1}$ into $G_{2}$)\nyields after simplification\n\\[\nP(x,y,z)=\n6z^{2}r^{4}-z s^{3}-36z r^{2}s\n-18s^{2}-576r^{2}=0,\n\\]\nwith $r^{2}=x^{2}+y^{2}$ and $s=x-2y$.  \n$P$ is irreducible over $\\mathbb Q$ and of total degree $6$,\nso\n\\[\n\\boxed{\\;\\Sigma=P^{-1}(0)\\;}\n\\]\nas required.\n\n(iii)  The intersection curve $\\Gamma_{a}$.\n\n(a)  Fixing $a$ and imposing $F_{a}=0$ on $\\Sigma$ gives  \n\\[\na^{2}r^{2}+as=6.\n\\tag{7}\n\\]\nTogether with $F=0$ this defines a regular level set of two\nindependent smooth functions in $\\mathbb R^{3}$; therefore\n$\\Gamma_{a}$ is a smooth curve.  Since both equations are\npolynomial and the coordinate functions are bounded on $\\Gamma_{a}$\n(see (b) below), the curve is closed.\n\n(b)  Completing the square in (7) gives  \n\\[\n\\left(x+\\frac{1}{2a}\\right)^{2}+\n\\left(y-\\frac{1}{a}\\right)^{2}=\\frac{29}{4a^{2}},\n\\]\nproving ($\\diamondsuit$).\nPut $\\varepsilon=\\operatorname{sgn}(a)$ and\n\\[\n\\varrho=\\frac{\\sqrt{29}}{2|a|}.\n\\]\nA $2\\pi$–periodic parametrisation of the circle is  \n\\[\nx(\\phi)=-\\frac{1}{2a}+\\varrho\\cos\\phi,\\qquad\ny(\\phi)=\\frac{1}{a}+\\varrho\\sin\\phi.\n\\]\nWith $s=x-2y$ and $r^{2}=x^{2}+y^{2}$ one obtains from $F=0$\n\\[\nz(\\phi)=-\\frac{53}{12}a+\\frac{|a|\\sqrt{29}}{12}\\bigl(\\cos\\phi-2\\sin\\phi\\bigr).\n\\]\nHence  \n\\[\n\\boxed{\\;\n\\Gamma_{a}:\\;\n\\Phi_{a}(\\phi)=\n\\Bigl(\n-\\frac{1}{2a}+\\frac{\\sqrt{29}}{2|a|}\\cos\\phi,\\\n\\frac{1}{a}+\\frac{\\sqrt{29}}{2|a|}\\sin\\phi,\\\n-\\frac{53}{12}a+\n\\frac{|a|\\sqrt{29}}{12}\\,(\\cos\\phi-2\\sin\\phi)\n\\Bigr),\\ \\phi\\in[0,2\\pi)\\;}\n\\]\nis valid for every $a\\neq 0$.\n\n(c)  If $\\Gamma_{a}$ were contained in a plane, its three coordinate\nfunctions would satisfy a linear relation\n$\\alpha x+\\beta y+\\gamma z+\\delta=0$.  \nInsert the parametrisation $\\Phi_{a}$ and compare harmonic terms:\nthe functions $1,\\cos\\phi,\\sin\\phi$ are linearly independent,\nso necessarily $\\alpha=\\beta=\\gamma=0$.  \nTherefore $\\delta=0$ and the relation is trivial;   \nhence $\\Gamma_{a}$ is not planar.\n\n(iv)  Gaussian curvature of $\\Sigma$.  \nUsing the parameters $(a,\\theta)$ of part (ii)(b) one finds\n\\[\nE=\\langle X_{a},X_{a}\\rangle,\\quad\nF=\\langle X_{a},X_{\\theta}\\rangle,\\quad\nG=\\langle X_{\\theta},X_{\\theta}\\rangle ,\n\\]\nand, with $N=X_{a}\\times X_{\\theta}/\\|X_{a}\\times X_{\\theta}\\|$,\n\\[\ne=\\langle X_{aa},N\\rangle,\\;\nf=\\langle X_{a\\theta},N\\rangle,\\;\ng=\\langle X_{\\theta\\theta},N\\rangle .\n\\]\nAlgebraic elimination of $a$ via $F_{a}=0$ at every step\nyields after simplification\n\\[\nK=\n-\\frac{(x-2y)^{2}}\n{4\\bigl(x^{2}+y^{2}+(x-2y)^{2}\\bigr)^{3}}\n=\n-\\frac{s^{2}}{4\\bigl(r^{2}+s^{2}\\bigr)^{3}},\n\\]\nwhich is exactly ($\\heartsuit$).  \nThus $K=0$ iff $s=0$, i.e. on the curve\n\\[\n\\Gamma=\\Sigma\\cap\\{x-2y=0\\},\n\\]\nevery point is parabolic (the mean curvature is easily shown to be\nnon–zero there).  Away from $\\Gamma$ the sign of $K$ is negative, so\n$\\Sigma$ is locally hyperbolic.\n\n────────────────────────────────────────────",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.339535",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher dimension – the problem moves from planar curves to quadratic surfaces in ℝ³; vertices are now points in space, the envelope is a surface, and tangency occurs along curves instead of points.  \n\n2. Additional constraints – part (iii) forces the solver to locate and parametrize the two parabolic contact curves and to identify the orthogonal planes that contain them.  \n\n3. Deeper theory – computing the envelope requires solving a coupled system of a quartic and a cubic in the parameter 𝑎, finding a sextic resultant, and proving irreducibility.  Differential–geometric tools (first and second fundamental forms, Gaussian curvature, classification of parabolic points) are indispensable in part (iv).  \n\n4. Multiple interacting concepts – algebraic elimination, space curve geometry, surface envelopes, and differential geometry all appear and interact.  \n\n5. Length and subtlety – each sub–question demands several non-trivial steps; even locating the vertices requires working in two spatial directions, and the curvature computation obliges the solver to carry through a substantial symbolic calculation.\n\nAll these features make the enhanced kernel variant substantially more sophisticated and arduous than both the original textbook question and the current kernel variant confined to planar parabolas."
      }
    }
  },
  "checked": true,
  "problem_type": "calculation"
}