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        "text_description": "Core Idea",
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        "text_description": "Ellipse\nAll points \\(P\\) satisfying \\(PF_{1}+PF_{2}=k\\), with fixed foci \\(F_{1},F_{2}\\) and constant \\(k > F_{1}F_{2}\\).\nKey Characteristics:\nTwo special points called foci guide the curve.\nSum of distances to the foci is constant for every point.\nThis constant exceeds the focal separation.\nExample:\nIf the sum equalled the focal distance, what locus would form?",
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  {
    "slide": 2,
    "fragments": [
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        "fragment_index": -1,
        "text_description": "Building the Equation\nFollow the algebraic trail from definition to equation.",
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        "text_description": "1\nSet the Coordinates\nPlace foci at \\(F_1(-c,0)\\) and \\(F_2(c,0)\\). Pick point \\(P(x,y)\\). Centre at origin, major axis on \\(x\\)-axis.",
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        "fragment_index": 2,
        "text_description": "2\nApply Distance Formula\nCompute \\(PF_1=\\sqrt{(x+c)^2+y^2}\\) and \\(PF_2=\\sqrt{(x-c)^2+y^2}\\). By definition, \\(PF_1+PF_2=2a\\).",
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        "fragment_index": 3,
        "text_description": "3\nIsolate \\(x\\) and \\(y\\)\nSquare twice and simplify to get \\( \\frac{x^2}{a^2} + \\frac{y^2}{b^2} = 1 \\) where \\(b^2 = a^2 - c^2\\).",
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      {
        "fragment_index": 4,
        "text_description": "Pro Tip:\nRemember \\(a \\gt b\\). If \\(c=0\\), the ellipse collapses into a circle.",
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  },
  {
    "slide": 3,
    "fragments": [
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        "fragment_index": -1,
        "text_description": "Crunching Numbers",
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        "text_description": "1\n\\[(x+c)^2 + y^2 + (x-c)^2 + y^2 = (2a)^2\\]\nSquare each focus distance and add; constant \\(2a\\) fixes ellipse size.",
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      {
        "fragment_index": 2,
        "text_description": "2\n\\[2x^2 + 2y^2 = 4a^2 - 2c^2\\]\nExpand brackets, then isolate \\(x\\) and \\(y\\) terms; constants shift right.",
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      {
        "fragment_index": 3,
        "text_description": "3\n\\[\\frac{x^2}{a^2} + \\frac{y^2}{b^2} = 1,\\; b^2 = a^2 - c^2\\]\nDivide by \\(a^2\\) and set \\(b\\); standard ellipse form emerges.",
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      {
        "fragment_index": 4,
        "text_description": "Key Insight:\nEach algebra step mirrors geometry: squaring fixes focus distances, isolation separates variables, \\(a\\) and \\(b\\) reveal semi-axes.",
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    ]
  },
  {
    "slide": 4,
    "fragments": [
      {
        "fragment_index": -1,
        "text_description": "Axis Relations",
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      {
        "fragment_index": 1,
        "text_description": "\\[a^{2}=b^{2}+c^{2}\\]\nRelation linking semi-major \\(a\\), semi-minor \\(b\\), and focal distance \\(c\\).",
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      {
        "fragment_index": 2,
        "text_description": "Variable Definitions\na\nSemi-major axis (longest radius)\nb\nSemi-minor axis (shorter radius)\nc\nDistance from centre \\(O\\) to each focus",
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      },
      {
        "fragment_index": 3,
        "text_description": "Applications\nFind Foci\nCompute \\(c=\\sqrt{a^{2}-b^{2}}\\) when axes lengths are known.\nCheck Axis Data\nVerify if three given lengths can form an ellipse by testing \\(a^{2}=b^{2}+c^{2}\\).\nDistance Proofs\nSupports proof that any point \\(P\\) satisfies \\(PF_{1}+PF_{2}=2a\\).",
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    ]
  },
  {
    "slide": 5,
    "fragments": [
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        "fragment_index": -1,
        "text_description": "Eccentricity e",
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      {
        "fragment_index": 1,
        "text_description": "\\(e = \\frac{c}{a}\\)",
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      },
      {
        "fragment_index": 2,
        "text_description": "Eccentricity \\(e\\) quantifies ellipse flatness: \\(0 < e < 1\\). \\(e\\) near 0 ⇒ almost circle; \\(e\\) near 1 ⇒ very elongated.",
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      {
        "fragment_index": 3,
        "text_description": "Earth’s orbit has \\(e \\approx 0.017\\) — practically circular.",
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    ]
  },
  {
    "slide": 6,
    "fragments": []
  },
  {
    "slide": 7,
    "fragments": [
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        "fragment_index": -1,
        "text_description": "Latus Rectum",
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        "text_description": "What is the Latus Rectum?\nThe latus rectum is a special chord of an ellipse.\nIt passes through a focus and is perpendicular to the major axis.\nKey Points:\nDefinition: chord through a focus \\(F(c,0)\\) ⟂ major axis.\nFind its ends by substituting \\(x=c\\) into \\( \\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1 \\).\nLength \\( \\text{LR}=2|y|=\\frac{2b^{2}}{a} \\).",
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  },
  {
    "slide": 8,
    "fragments": [
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        "fragment_index": -1,
        "text_description": "Elliptical Takeaways\nKey facts in one glance",
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        "text_description": "Definition\nSet of all points whose distances to two foci add to \\(2a\\).",
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      {
        "fragment_index": 2,
        "text_description": "Standard Equation\n\\(\\displaystyle \\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1,\\; a>b>0\\).",
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        "text_description": "Focus Relation\nSemi-axes satisfy \\(a^{2}=b^{2}+c^{2}\\).",
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      {
        "fragment_index": 4,
        "text_description": "Eccentricity\n\\(e=\\frac{c}{a},\\; 0<e<1\\) gauges “ovalness”.",
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      {
        "fragment_index": 5,
        "text_description": "Latus Rectum\nThrough a focus, length \\( \\frac{2b^{2}}{a}\\).",
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      }
    ]
  }
]