{
  "index": "1941-B-4",
  "type": "GEO",
  "tag": [
    "GEO",
    "ANA"
  ],
  "difficulty": "",
  "question": "11. Two perpendicular diameters of the ellipse\n\\[\n\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1\n\\]\nare given, and the two diameters conjugate to them are constructed. Show that the rectangular hyperbola passing through the ends of these conjugate diameters passes through the foci of the ellipse.",
  "solution": "Solution. Let \\( E \\) be the given ellipse. We assume that \\( a^{2}>b^{2} \\). Then the foci are at \\( (c, 0) \\) and \\( (-c, 0) \\), where \\( c=\\sqrt{a^{2}-b^{2}} \\).\n\nSuppose the given diameters have equations \\( y=m x \\) and \\( -m y=x \\). Recall that one diameter of an ellipse is conjugate to another if it is parallel to the tangents at the ends of the other. The conjugate diameters are easily found to have the equations\n\\[\nb^{2} x+m a^{2} y=0 \\text { and } m b^{2} x-a^{2} y=0\n\\]\nrespectively. Hence\n\\[\n\\left(b^{2} x+m a^{2} y\\right)\\left(m b^{2} x-a^{2} y\\right)=m b^{4} x^{2}-m a^{4} y^{2}+\\left(m^{2}-1\\right) a^{2} b^{2} x y=0\n\\]\nis the equation of the degenerate conic \\( D \\) consisting of the lines of the two conjugate diameters. Any conic (except \\( E \\) ) passing through the four points where \\( D \\) meets \\( E \\) (i.e., the ends of the conjugate diameters) has an equation of the form\n\\[\n\\lambda\\left(\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}-1\\right)+m b^{4} x^{2}-m a^{4} y^{2}+\\left(m^{2}-1\\right) a^{2} b^{2} x y=0 .\n\\]\n\nThis conic is a rectangular hyperbola (possibly degenerate) if and only if the sum of the coefficients of \\( x^{2} \\) and \\( y^{2} \\) is zero, that is,\n\\[\n\\lambda\\left(\\frac{1}{a^{2}}+\\frac{1}{b^{2}}\\right)+m b^{4}-m a^{4}=0,\n\\]\ni.e., \\( \\lambda=m a^{2} b^{2} c^{2} \\). With this value of \\( \\lambda \\), the conic (1) passes through \\( (c, 0) \\) and \\( (-c, 0) \\), as required.\n\nThe rectangular hyperbola, (1) with \\( \\lambda=m a^{2} b^{2} c^{2} \\), is degenerate if and only if the constant term \\( \\lambda \\) is zero. If the given ellipse is a genuine ellipse (i.e., \\( a^{2} \\neq b^{2} \\) ), this occurs only for \\( m=0 \\), that is, when the original diameters are the axes of the ellipse. In this case the conjugate diameters are also the axes and there is no proper rectangular hyperbola through the four points. The union of the two axes is then a degenerate rectangular hyperbola fulfilling the conditions.\n\nIf the given ellipse is actually a circle, \\( c=0 \\), so \\( \\lambda \\) is always zero. For a circle, perpendicular diameters are always conjugate and the union of any two such diameters is a degenerate rectangular hyperbola which passes through the foci, which coincide at the center.",
  "vars": [
    "x",
    "y",
    "E",
    "D"
  ],
  "params": [
    "a",
    "b",
    "c",
    "m",
    "\\\\lambda"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "abscissa",
        "y": "ordinate",
        "E": "ellcurve",
        "D": "degencon",
        "a": "semimajor",
        "b": "semiminor",
        "c": "focusdist",
        "m": "slopeval",
        "\\lambda": "lambdavar"
      },
      "question": "11. Two perpendicular diameters of the ellipse\n\\[\n\\frac{abscissa^{2}}{semimajor^{2}}+\\frac{ordinate^{2}}{semiminor^{2}}=1\n\\]\nare given, and the two diameters conjugate to them are constructed. Show that the rectangular hyperbola passing through the ends of these conjugate diameters passes through the foci of the ellipse.",
      "solution": "Solution. Let \\( ellcurve \\) be the given ellipse. We assume that \\( semimajor^{2}>semiminor^{2} \\). Then the foci are at \\( (focusdist, 0) \\) and \\( (-focusdist, 0) \\), where \\( focusdist=\\sqrt{semimajor^{2}-semiminor^{2}} \\).\n\nSuppose the given diameters have equations \\( ordinate=slopeval\\, abscissa \\) and \\( -slopeval\\, ordinate=abscissa \\). Recall that one diameter of an ellipse is conjugate to another if it is parallel to the tangents at the ends of the other. The conjugate diameters are easily found to have the equations\n\\[\nsemiminor^{2}\\, abscissa + slopeval\\, semimajor^{2}\\, ordinate = 0 \\text { and } slopeval\\, semiminor^{2}\\, abscissa - semimajor^{2}\\, ordinate = 0\n\\]\nrespectively. Hence\n\\[\n\\left(semiminor^{2}\\, abscissa + slopeval\\, semimajor^{2}\\, ordinate\\right)\\left(slopeval\\, semiminor^{2}\\, abscissa - semimajor^{2}\\, ordinate\\right)=slopeval\\, semiminor^{4}\\, abscissa^{2}-slopeval\\, semimajor^{4}\\, ordinate^{2}+\\left(slopeval^{2}-1\\right) semimajor^{2} semiminor^{2}\\, abscissa\\, ordinate = 0\n\\]\nis the equation of the degenerate conic \\( degencon \\) consisting of the lines of the two conjugate diameters. Any conic (except \\( ellcurve \\) ) passing through the four points where \\( degencon \\) meets \\( ellcurve \\) (i.e., the ends of the conjugate diameters) has an equation of the form\n\\[\nlambdavar\\left(\\frac{abscissa^{2}}{semimajor^{2}}+\\frac{ordinate^{2}}{semiminor^{2}}-1\\right)+slopeval\\, semiminor^{4}\\, abscissa^{2}-slopeval\\, semimajor^{4}\\, ordinate^{2}+\\left(slopeval^{2}-1\\right) semimajor^{2} semiminor^{2}\\, abscissa\\, ordinate = 0 .\n\\]\n\nThis conic is a rectangular hyperbola (possibly degenerate) if and only if the sum of the coefficients of \\( abscissa^{2} \\) and \\( ordinate^{2} \\) is zero, that is,\n\\[\nlambdavar\\left(\\frac{1}{semimajor^{2}}+\\frac{1}{semiminor^{2}}\\right)+slopeval\\, semiminor^{4}-slopeval\\, semimajor^{4}=0,\n\\]\ni.e., \\( lambdavar=slopeval\\, semimajor^{2} semiminor^{2} focusdist^{2} \\). With this value of \\( lambdavar \\), the conic (1) passes through \\( (focusdist, 0) \\) and \\( (-focusdist, 0) \\), as required.\n\nThe rectangular hyperbola, (1) with \\( lambdavar=slopeval\\, semimajor^{2} semiminor^{2} focusdist^{2} \\), is degenerate if and only if the constant term \\( lambdavar \\) is zero. If the given ellipse is a genuine ellipse (i.e., \\( semimajor^{2} \\neq semiminor^{2} \\) ), this occurs only for \\( slopeval=0 \\), that is, when the original diameters are the axes of the ellipse. In this case the conjugate diameters are also the axes and there is no proper rectangular hyperbola through the four points. The union of the two axes is then a degenerate rectangular hyperbola fulfilling the conditions.\n\nIf the given ellipse is actually a circle, \\( focusdist=0 \\), so \\( lambdavar \\) is always zero. For a circle, perpendicular diameters are always conjugate and the union of any two such diameters is a degenerate rectangular hyperbola which passes through the foci, which coincide at the center."
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "pineconee",
        "y": "snowflaker",
        "E": "riverdelta",
        "D": "cliffhanger",
        "a": "cornerstone",
        "b": "windsurfer",
        "c": "marshmallow",
        "m": "afterglow",
        "\\lambda": "honeybadger"
      },
      "question": "11. Two perpendicular diameters of the ellipse\n\\[\n\\frac{pineconee^{2}}{cornerstone^{2}}+\\frac{snowflaker^{2}}{windsurfer^{2}}=1\n\\]\nare given, and the two diameters conjugate to them are constructed. Show that the rectangular hyperbola passing through the ends of these conjugate diameters passes through the foci of the ellipse.",
      "solution": "Solution. Let \\( riverdelta \\) be the given ellipse. We assume that \\( cornerstone^{2}>windsurfer^{2} \\). Then the foci are at \\( (marshmallow, 0) \\) and \\( (-marshmallow, 0) \\), where \\( marshmallow=\\sqrt{cornerstone^{2}-windsurfer^{2}} \\).\n\nSuppose the given diameters have equations \\( snowflaker=afterglow\\,pineconee \\) and \\( -afterglow\\,snowflaker=pineconee \\). Recall that one diameter of an ellipse is conjugate to another if it is parallel to the tangents at the ends of the other. The conjugate diameters are easily found to have the equations\n\\[\nwindsurfer^{2} pineconee+afterglow cornerstone^{2} snowflaker=0 \\text { and } afterglow windsurfer^{2} pineconee-cornerstone^{2} snowflaker=0\n\\]\nrespectively. Hence\n\\[\n\\left(windsurfer^{2} pineconee+afterglow cornerstone^{2} snowflaker\\right)\\left(afterglow windsurfer^{2} pineconee-cornerstone^{2} snowflaker\\right)=afterglow windsurer^{4} pineconee^{2}-afterglow cornerstone^{4} snowflaker^{2}+\\left(afterglow^{2}-1\\right) cornerstone^{2} windsurer^{2} pineconee snowflaker=0\n\\]\nis the equation of the degenerate conic \\( cliffhanger \\) consisting of the lines of the two conjugate diameters. Any conic (except \\( riverdelta \\) ) passing through the four points where \\( cliffhanger \\) meets \\( riverdelta \\) (i.e., the ends of the conjugate diameters) has an equation of the form\n\\[\nhoneybadger\\left(\\frac{pineconee^{2}}{cornerstone^{2}}+\\frac{snowflaker^{2}}{windsurfer^{2}}-1\\right)+afterglow windsurer^{4} pineconee^{2}-afterglow cornerstone^{4} snowflaker^{2}+\\left(afterglow^{2}-1\\right) cornerstone^{2} windsurer^{2} pineconee snowflaker=0 .\n\\]\n\nThis conic is a rectangular hyperbola (possibly degenerate) if and only if the sum of the coefficients of \\( pineconee^{2} \\) and \\( snowflaker^{2} \\) is zero, that is,\n\\[\nhoneybadger\\left(\\frac{1}{cornerstone^{2}}+\\frac{1}{windsurfer^{2}}\\right)+afterglow windsurer^{4}-afterglow cornerstone^{4}=0,\n\\]\ni.e., \\( honeybadger=afterglow cornerstone^{2} windsurer^{2} marshmallow^{2} \\). With this value of \\( honeybadger \\), the conic (1) passes through \\( (marshmallow, 0) \\) and \\( (-marshmallow, 0) \\), as required.\n\nThe rectangular hyperbola, (1) with \\( honeybadger=afterglow cornerstone^{2} windsurer^{2} marshmallow^{2} \\), is degenerate if and only if the constant term \\( honeybadger \\) is zero. If the given ellipse is a genuine ellipse (i.e., \\( cornerstone^{2} \\neq windsurer^{2} \\) ), this occurs only for \\( afterglow=0 \\), that is, when the original diameters are the axes of the ellipse. In this case the conjugate diameters are also the axes and there is no proper rectangular hyperbola through the four points. The union of the two axes is then a degenerate rectangular hyperbola fulfilling the conditions.\n\nIf the given ellipse is actually a circle, \\( marshmallow=0 \\), so \\( honeybadger \\) is always zero. For a circle, perpendicular diameters are always conjugate and the union of any two such diameters is a degenerate rectangular hyperbola which passes through the foci, which coincide at the center."
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "verticalordinate",
        "y": "horizontalabscissa",
        "E": "rectangleshape",
        "D": "fullcircle",
        "a": "minorbasis",
        "b": "majorbasis",
        "c": "nonfocal",
        "m": "flatness",
        "\\lambda": "fixedness"
      },
      "question": "11. Two perpendicular diameters of the ellipse\n\\[\n\\frac{verticalordinate^{2}}{minorbasis^{2}}+\\frac{horizontalabscissa^{2}}{majorbasis^{2}}=1\n\\]\nare given, and the two diameters conjugate to them are constructed. Show that the rectangular hyperbola passing through the ends of these conjugate diameters passes through the foci of the ellipse.",
      "solution": "Solution. Let \\( rectangleshape \\) be the given ellipse. We assume that \\( minorbasis^{2}>majorbasis^{2} \\). Then the foci are at \\( (nonfocal, 0) \\) and \\( (-nonfocal, 0) \\), where \\( nonfocal=\\sqrt{minorbasis^{2}-majorbasis^{2}} \\).\n\nSuppose the given diameters have equations \\( horizontalabscissa=flatness\\,verticalordinate \\) and \\( -flatness\\,horizontalabscissa=verticalordinate \\). Recall that one diameter of an ellipse is conjugate to another if it is parallel to the tangents at the ends of the other. The conjugate diameters are easily found to have the equations\n\\[\nmajorbasis^{2} verticalordinate+flatness minorbasis^{2} horizontalabscissa=0 \\text { and } flatness majorbasis^{2} verticalordinate- minorbasis^{2} horizontalabscissa=0\n\\]\nrespectively. Hence\n\\[\n\\left(majorbasis^{2} verticalordinate+flatness minorbasis^{2} horizontalabscissa\\right)\\left(flatness majorbasis^{2} verticalordinate- minorbasis^{2} horizontalabscissa\\right)=flatness majorbasis^{4} verticalordinate^{2}-flatness minorbasis^{4} horizontalabscissa^{2}+\\left(flatness^{2}-1\\right) minorbasis^{2} majorbasis^{2} verticalordinate horizontalabscissa=0\n\\]\nis the equation of the degenerate conic \\( fullcircle \\) consisting of the lines of the two conjugate diameters. Any conic (except \\( rectangleshape \\) ) passing through the four points where \\( fullcircle \\) meets \\( rectangleshape \\) (i.e., the ends of the conjugate diameters) has an equation of the form\n\\[\nfixedness\\left(\\frac{verticalordinate^{2}}{minorbasis^{2}}+\\frac{horizontalabscissa^{2}}{majorbasis^{2}}-1\\right)+flatness majorbasis^{4} verticalordinate^{2}-flatness minorbasis^{4} horizontalabscissa^{2}+\\left(flatness^{2}-1\\right) minorbasis^{2} majorbasis^{2} verticalordinate horizontalabscissa=0 .\n\\]\n\nThis conic is a rectangular hyperbola (possibly degenerate) if and only if the sum of the coefficients of \\( verticalordinate^{2} \\) and \\( horizontalabscissa^{2} \\) is zero, that is,\n\\[\nfixedness\\left(\\frac{1}{minorbasis^{2}}+\\frac{1}{majorbasis^{2}}\\right)+flatness majorbasis^{4}-flatness minorbasis^{4}=0,\n\\]\ni.e., \\( fixedness=flatness minorbasis^{2} majorbasis^{2} nonfocal^{2} \\). With this value of \\( fixedness \\), the conic (1) passes through \\( (nonfocal, 0) \\) and \\( (-nonfocal, 0) \\), as required.\n\nThe rectangular hyperbola, (1) with \\( fixedness=flatness minorbasis^{2} majorbasis^{2} nonfocal^{2} \\), is degenerate if and only if the constant term \\( fixedness \\) is zero. If the given ellipse is a genuine ellipse (i.e., \\( minorbasis^{2} \\neq majorbasis^{2} \\) ), this occurs only for \\( flatness=0 \\), that is, when the original diameters are the axes of the ellipse. In this case the conjugate diameters are also the axes and there is no proper rectangular hyperbola through the four points. The union of the two axes is then a degenerate rectangular hyperbola fulfilling the conditions.\n\nIf the given ellipse is actually a circle, \\( nonfocal=0 \\), so \\( fixedness \\) is always zero. For a circle, perpendicular diameters are always conjugate and the union of any two such diameters is a degenerate rectangular hyperbola which passes through the foci, which coincide at the center."
    },
    "garbled_string": {
      "map": {
        "x": "qzxwvtnp",
        "y": "hjgrksla",
        "E": "vbmncrtle",
        "D": "pfrqslzne",
        "a": "lkjhgfdsw",
        "b": "nmzxcvbal",
        "c": "wertyuiop",
        "m": "sdfghjklq",
        "\\lambda": "poilkjmnb"
      },
      "question": "11. Two perpendicular diameters of the ellipse\n\\[\n\\frac{qzxwvtnp^{2}}{lkjhgfdsw^{2}}+\\frac{hjgrksla^{2}}{nmzxcvbal^{2}}=1\n\\]\nare given, and the two diameters conjugate to them are constructed. Show that the rectangular hyperbola passing through the ends of these conjugate diameters passes through the foci of the ellipse.",
      "solution": "Solution. Let \\( vbmncrtle \\) be the given ellipse. We assume that \\( lkjhgfdsw^{2}>nmzxcvbal^{2} \\). Then the foci are at \\( (wertyuiop, 0) \\) and \\( (-wertyuiop, 0) \\), where \\( wertyuiop=\\sqrt{lkjhgfdsw^{2}-nmzxcvbal^{2}} \\).\n\nSuppose the given diameters have equations \\( hjgrksla=sdfghjklq qzxwvtnp \\) and \\( -sdfghjklq hjgrksla=qzxwvtnp \\). Recall that one diameter of an ellipse is conjugate to another if it is parallel to the tangents at the ends of the other. The conjugate diameters are easily found to have the equations\n\\[\nnmzxcvbal^{2} qzxwvtnp+sdfghjklq lkjhgfdsw^{2} hjgrksla=0 \\text { and } sdfghjklq nmzxcvbal^{2} qzxwvtnp-lkjhgfdsw^{2} hjgrksla=0\n\\]\nrespectively. Hence\n\\[\n\\left(nmzxcvbal^{2} qzxwvtnp+sdfghjklq lkjhgfdsw^{2} hjgrksla\\right)\\left(sdfghjklq nmzxcvbal^{2} qzxwvtnp-lkjhgfdsw^{2} hjgrksla\\right)=sdfghjklq nmzxcvbal^{4} qzxwvtnp^{2}-sdfghjklq lkjhgfdsw^{4} hjgrksla^{2}+\\left(sdfghjklq^{2}-1\\right) lkjhgfdsw^{2} nmzxcvbal^{2} qzxwvtnp hjgrksla=0\n\\]\nis the equation of the degenerate conic \\( pfrqslzne \\) consisting of the lines of the two conjugate diameters. Any conic (except \\( vbmncrtle \\) ) passing through the four points where \\( pfrqslzne \\) meets \\( vbmncrtle \\) (i.e., the ends of the conjugate diameters) has an equation of the form\n\\[\npoilkjmnb\\left(\\frac{qzxwvtnp^{2}}{lkjhgfdsw^{2}}+\\frac{hjgrksla^{2}}{nmzxcvbal^{2}}-1\\right)+sdfghjklq nmzxcvbal^{4} qzxwvtnp^{2}-sdfghjklq lkjhgfdsw^{4} hjgrksla^{2}+\\left(sdfghjklq^{2}-1\\right) lkjhgfdsw^{2} nmzxcvbal^{2} qzxwvtnp hjgrksla=0 .\n\\]\n\nThis conic is a rectangular hyperbola (possibly degenerate) if and only if the sum of the coefficients of \\( qzxwvtnp^{2} \\) and \\( hjgrksla^{2} \\) is zero, that is,\n\\[\npoilkjmnb\\left(\\frac{1}{lkjhgfdsw^{2}}+\\frac{1}{nmzxcvbal^{2}}\\right)+sdfghjklq nmzxcvbal^{4}-sdfghjklq lkjhgfdsw^{4}=0,\n\\]\ni.e., \\( poilkjmnb=sdfghjklq lkjhgfdsw^{2} nmzxcvbal^{2} wertyuiop^{2} \\). With this value of \\( poilkjmnb \\), the conic (1) passes through \\( (wertyuiop, 0) \\) and \\( (-wertyuiop, 0) \\), as required.\n\nThe rectangular hyperbola, (1) with \\( poilkjmnb=sdfghjklq lkjhgfdsw^{2} nmzxcvbal^{2} wertyuiop^{2} \\), is degenerate if and only if the constant term \\( poilkjmnb \\) is zero. If the given ellipse is a genuine ellipse (i.e., \\( lkjhgfdsw^{2} \\neq nmzxcvbal^{2} \\) ), this occurs only for \\( sdfghjklq=0 \\), that is, when the original diameters are the axes of the ellipse. In this case the conjugate diameters are also the axes and there is no proper rectangular hyperbola through the four points. The union of the two axes is then a degenerate rectangular hyperbola fulfilling the conditions.\n\nIf the given ellipse is actually a circle, \\( wertyuiop=0 \\), so \\( poilkjmnb \\) is always zero. For a circle, perpendicular diameters are always conjugate and the union of any two such diameters is a degenerate rectangular hyperbola which passes through the foci, which coincide at the center."
    },
    "kernel_variant": {
      "question": "Let 𝔈 be the ellipse\n\\[\n\\frac{x^{2}}{\\rho^{2}}+\\frac{y^{2}}{\\sigma^{2}}=1\\qquad(0<\\rho<\\sigma),\n\\]\nso that its two foci are F₁=(0,\\;d) and F₂=(0,\\;-d), where d^{2}=\\sigma^{2}-\\rho^{2}.  \nThrough the centre O we are given the two perpendicular diameters\n\\[\ny=kx\\quad\\text{and}\\quad y=-\\frac{x}{k}\\qquad(k\\ne0).\n\\]\nFor each of these diameters construct the diameter conjugate to it (that is, the line through O parallel to the tangents at the ends of the given diameter).  \nShow that the unique rectangular hyperbola that passes through the four endpoints of the two conjugate diameters also passes through the two foci F₁ and F₂ of the ellipse.",
      "solution": "Let\n            E :   x²/ρ² + y²/σ² = 1     (0 < ρ < σ,   d² = σ² – ρ²).\nTwo perpendicular diameters issuing from the centre O are\n            y = kx   and   y = –x/k   (k ≠ 0).\nWe show that the rectangular hyperbola through the ends of the conjugates of these diameters contains the foci F₁(0, d) and F₂(0, –d).\n\nStep 1  (conjugate diameters).\nFor P(x₀,y₀) on E the tangent is\n            x x₀/ρ² + y y₀/σ² = 1.                (★)\nIf P lies on y = kx, then y₀ = kx₀ and (★) becomes\n            (x/ρ²) + k(y/σ²) = 1/x₀.             (1)\nThe normal vector of this tangent is therefore proportional to\n            (1, k ρ²/σ²).\nHence the line through O that is parallel to the tangent, i.e. the diameter conjugate to y = kx, is\n            σ² x + k ρ² y = 0.                         (C₁)\nRepeating the computation for the diameter y = –x/k gives the conjugate\n            k σ² x – ρ² y = 0.                         (C₂)\n\nStep 2  (degenerate conic through the four constructed points).\nThe two lines (C₁) and (C₂) form the degenerate conic\n            (σ²x + kρ²y)(kσ²x – ρ²y) = 0\n⇔   k σ⁴ x² – k ρ⁴ y² + (k² – 1) ρ²σ² xy = 0.          (D)\n\nStep 3  (the pencil of conics through the four points).\nEvery conic through the four intersection points E ∩ D is a member of the pencil spanned by E and D, i.e.\n      C(λ):   λ(x²/ρ² + y²/σ² – 1)\n              + kσ⁴x² – kρ⁴y² + (k² – 1)ρ²σ² xy = 0          (2)\nwhere λ ∈ ℝ.\n\nStep 4  (rectangular–hyperbola condition).\nFor a central conic Ax² + 2Hxy + By² + … = 0 the axes are perpendicular (the conic is rectangular) iff A + B = 0.  In (2)\n      A = λ/ρ² + kσ⁴,     B = λ/σ² – kρ⁴,\nso the condition A + B = 0 gives\n      λ(1/ρ² + 1/σ²) + k(σ⁴ – ρ⁴) = 0\n⇔   λ = –k ρ² σ² (σ² – ρ²) = –k ρ² σ² d².                (3)\nFor k ≠ 0 this indeed yields a non‐degenerate rectangular hyperbola.\n\nStep 5  (the hyperbola contains the foci).\nInsert x = 0, y = d into (2) together with (3).  Only two terms survive:\n      λ(d²/σ² – 1) – kρ⁴ d² = 0.\nBecause d² = σ² – ρ², the bracket equals –ρ²/σ², and with (3)\n      [–kρ²σ²d²](–ρ²/σ²) – kρ⁴d² = kρ⁴d² – kρ⁴d² = 0.\nThus (0,d) lies on the conic, and the same calculation with y = –d shows that (0,–d) does as well.\n\nConsequently the unique rectangular hyperbola through the four endpoints of the two conjugate diameters of the ellipse passes through the foci F₁ and F₂.",
      "_meta": {
        "core_steps": [
          "Describe the two perpendicular diameters (slope m and −1/m) and compute their conjugate diameters using the tangent-parallel definition.",
          "Combine the two conjugate‐diameter lines into a degenerate conic L₁·L₂ = 0.",
          "Form the pencil λ·(ellipse) + (degenerate conic) = 0 that contains every conic through the four endpoints of the conjugate diameters.",
          "Impose the rectangular-hyperbola condition (sum of x²– and y²–coefficients equals 0) to determine λ.",
          "Substitute this λ to verify that the resulting conic passes through the ellipse’s foci, completing the proof."
        ],
        "mutable_slots": {
          "slot1": {
            "description": "Positive semi-axes of the ellipse (major/minor lengths). Any unequal pair keeps the argument intact.",
            "original": "a , b   in  x²/a² + y²/b² = 1"
          },
          "slot2": {
            "description": "Slope that fixes the orientation of the first given diameter; the second diameter is its perpendicular.",
            "original": "m   in  y = m x  and  x = −m y"
          },
          "slot3": {
            "description": "Choice of which axis is the major one; presently a² > b² so the foci are on the x-axis. Reversing roles (b > a) or rotating axes leaves the reasoning unchanged.",
            "original": "Assumption a² > b²  ⇒  foci at (±c, 0)"
          },
          "slot4": {
            "description": "Scalar parameter of the pencil of conics, later fixed by the rectangular-hyperbola condition.",
            "original": "λ   (becomes  λ = m a² b² c² )"
          }
        }
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}