{"id":806,"date":"2013-06-08T01:05:21","date_gmt":"2013-06-07T16:05:21","guid":{"rendered":"http:\/\/www.roundown.net\/nyushi\/?p=806"},"modified":"2021-09-24T17:42:21","modified_gmt":"2021-09-24T08:42:21","slug":"tok201305","status":"publish","type":"post","link":"https:\/\/www.roundown.net\/nyushi\/tok201305\/","title":{"rendered":"\u6771\u5de5\u59272013\uff1a\u7b2c5\u554f"},"content":{"rendered":"
\n

\\(a , b\\) \u3092\u6b63\u306e\u5b9f\u6570\u3068\u3057, \u5186 \\(C _ 1 : \\ (x-a)^2+y^2 = a^2\\) \u3068\u6955\u5186 \\(C _ 2 : \\ x^2+\\dfrac{y^2}{b^2} = 1\\) \u3092\u8003\u3048\u308b.<\/p>\r\n

    \r\n

  1. (1)<\/strong>\u3000\\(C _ 1\\) \u3068 \\(C _ 2\\) \u306b\u5185\u63a5\u3059\u308b\u305f\u3081\u306e \\(a , b\\) \u306e\u6761\u4ef6\u3092\u6c42\u3081\u3088.<\/p><\/li>\r\n

  2. (2)<\/strong>\u3000\\(b = \\dfrac{1}{\\sqrt{3}}\\) \u3068\u3057, \\(C _ 1\\) \u304c \\(C _ 2\\) \u306b\u5185\u63a5\u3057\u3066\u3044\u308b\u3068\u3059\u308b. \u3053\u306e\u3068\u304d, \u7b2c \\(1\\) \u8c61\u9650\u306b\u304a\u3051\u308b \\(C _ 1\\) \u3068 \\(C _ 2\\) \u306e\u63a5\u70b9\u306e\u5ea7\u6a19\u3092 \\((p,q)\\) \u3092\u6c42\u3081\u3088.<\/p><\/li>\r\n

  3. (3)<\/strong>\u3000(2)<\/strong> \u306e\u6761\u4ef6\u306e\u3082\u3068\u3067, \\(x \\geqq p\\) \u306e\u7bc4\u56f2\u306b\u304a\u3044\u3066, \\(C _ 1\\) \u3068 \\(C _ 2\\) \u3067\u56f2\u307e\u308c\u305f\u90e8\u5206\u306e\u9762\u7a4d\u3092\u6c42\u3081\u3088.<\/p><\/li>\r\n<\/ol>\r\n

    \r\n\r\n

    \u3010 \u89e3 \u7b54 \u3011<\/h2>\r\n

    (1)<\/strong><\/p>\r\n

    \u4e0b\u56f3\u306e\u3088\u3046\u306b, \u5186 \\(C _ 1\\) \u306f\u4e2d\u5fc3 \\((a,0)\\) , \u534a\u5f84 \\(a\\) \u3067\u3042\u308a, \u6955\u5186 \\(C _ 2\\) \u306f\u539f\u70b9\u4e2d\u5fc3\u3067\u70b9 \\((1,0)\\) \u3068 \\((0,b)\\) \u3092\u901a\u308b.<\/p>\r\n\"toko_2013_05_01\"\r\n

    \u3057\u305f\u304c\u3063\u3066, \\(x \\geqq 0 , \\ y \\geqq 0\\) \u306e\u9818\u57df\u306b\u304a\u3044\u3066, \\(C _ 1\\) \u304c \\(C _ 2\\) \u306b\u5185\u63a5\u3059\u308b\u6761\u4ef6\u3092\u8003\u3048\u308c\u3070\u3088\u3044.
    \r\n\\(C _ 2\\) \u4e0a\u306e\u70b9 P \\((p,q) \\ ( 0 \\leqq p \\leqq 1 )\\) \u3068 \\(C _ 1\\) \u306e\u4e2d\u5fc3 A \\((a,0)\\) \u3068\u306e\u8ddd\u96e2\u306e\u6700\u5c0f\u5024\u304c \\(a\\) \u306b\u306a\u308b\u3088\u3046\u306a \\(a , b\\) \u306e\u6761\u4ef6\u3092\u8003\u3048\u308c\u3070\u3088\u3044.\r\n\\[\\begin{align}\r\n\\text{AP}^2 & = (p-a)^2 +q^2 \\\\\r\n& = (p-a)^2+b^2(1-p^2) \\\\\r\n& = (1-b^2)p^2 -2ap +a^2+b^2 \\\\\r\n& = (1-b^2) \\left( p-\\dfrac{a}{1-b^2} \\right)^2 +a^2 +b^2 -\\dfrac{a^2}{1-b^2}\r\n\\end{align}\\]\r\n\u3053\u308c\u3092 \\(f(p)\\) \u3068\u304a\u304f.<\/p>\r\n

      \r\n

    1. 1*<\/strong>\u3000\\(b = 1\\) \u306e\u3068\u304d
      \r\n\u3053\u306e\u3068\u304d\r\n\\[\r\nf(p) = -2ap+a^2+1\r\n\\]\r\n\u3053\u308c\u306f \\(p\\) \u306e\u50be\u304d\u304c\u8ca0\u3067\u3042\u308b \\(1\\) \u6b21\u95a2\u6570\u306a\u306e\u3067, \u6700\u5c0f\u5024\u3068\u306a\u308a\u3046\u308b\u306e\u306f, \\(f(1)\\) \u3067\u3042\u308b.\r\n\\[\\begin{align}\r\nf(1) = a^2-2a+1 & = a^2 \\\\\r\n-2a+1 & = 0 \\\\\r\n\\text{\u2234} \\quad a & = \\dfrac{1}{2}\r\n\\end{align}\\]<\/li>\r\n

    2. 2*<\/strong>\u3000\\(b \\gt 1\\) \u306e\u3068\u304d
      \r\n\u3053\u306e\u3068\u304d, \\(f(p)\\) \u306f\u4e0a\u306b\u51f8\u306a \\(p\\) \u306e \\(2\\) \u6b21\u95a2\u6570\u3067\u3042\u308a, \u653e\u7269\u7dda\u306e\u8ef8\uff1a \\(p = \\dfrac{a}{1-b^2} \\lt 0\\) \u306a\u306e\u3067, \u6700\u5c0f\u5024\u306f \\(f(1)\\) \u3067\u3042\u308b.\r\n\\[\\begin{align}\r\nf(1) = a^2-2a+1 & = a^2 \\\\\r\n\\text{\u2234} \\quad a & = \\dfrac{1}{2}\r\n\\end{align}\\]<\/li>\r\n

    3. 3*<\/strong>\u3000\\(0 \\lt b \\lt 1\\) \u306e\u3068\u304d
      \r\n\u3053\u306e\u3068\u304d, \\(f(p)\\) \u306f\u4e0b\u306b\u51f8\u306a \\(p\\) \u306e \\(2\\) \u6b21\u95a2\u6570\u3067\u3042\u308a, \u653e\u7269\u7dda\u306e\u8ef8\uff1a \\(p = \\dfrac{a}{1-b^2} \\gt 0\\) \u306a\u306e\u3067,<\/p>\r\n

        \r\n
      1. (\u30a2)<\/strong>\u3000\\(0 \\lt \\dfrac{a}{1-b^2} \\lt 1\\) \u3059\u306a\u308f\u3061 \\(0 \\lt a \\lt 1-b^2\\) ...[1] \u306e\u3068\u304d
        \r\n\u6700\u5c0f\u5024\u306f \\(f \\left( \\dfrac{a}{1-b^2} \\right)\\) \u306a\u306e\u3067\r\n\\[\\begin{align}\r\nf \\left( \\dfrac{a}{1-b^2} \\right) = a^2 +b^2 & -\\dfrac{a^2}{1-b^2} = a^2 \\\\\r\na^2 & = b^2(1-b^2) \\\\\r\n\\text{\u2234} \\quad a & = b \\sqrt{1-b^2}\r\n\\end{align}\\]\r\n\u3053\u306e\u3068\u304d, [1] \u3088\u308a\r\n\\[\\begin{align}\r\n0 & \\lt b \\sqrt{1-b^2} \\lt 1-b^2 \\\\\r\n0 & \\lt b \\lt \\sqrt{1-b^2} \\\\\r\n\\text{\u2234} \\quad 0 & \\lt b^2 \\lt 1-b^2 \\\\\r\n\\text{\u2234} \\quad 0 & \\lt b \\lt \\dfrac{\\sqrt{2}}{2}\r\n\\end{align}\\]<\/li>\r\n
      2. (\u30a4)<\/strong>\u3000\\(\\dfrac{a}{1-b^2} \\geqq 1\\) \u3059\u306a\u308f\u3061 \\(a \\geqq 1-b^2\\) ...[2] \u306e\u3068\u304d
        \r\n\u6700\u5c0f\u5024\u306f \\(f(1)\\) \u306a\u306e\u3067\r\n\\[\\begin{align}\r\nf(1) = a^2-2a+1 & = a^2 \\\\\r\n\\text{\u2234} \\quad a & = \\dfrac{1}{2}\r\n\\end{align}\\]\r\n\u3053\u306e\u3068\u304d, [2] \u3088\u308a\r\n\\[\\begin{align}\r\n\\dfrac{1}{2} & \\geqq 1-b^2 \\\\\r\n\\text{\u2234} \\quad b & \\geqq \\dfrac{\\sqrt{2}}{2}\r\n\\end{align}\\]<\/li>\r\n<\/ol><\/li>\r\n<\/ol>\r\n

        \u4ee5\u4e0a\u3088\u308a, \u6c42\u3081\u308b\u6761\u4ef6\u306f\r\n\\[\r\n\\underline{\\left\\{\\begin{array}{ll} a = b \\sqrt{1-b^2} & \\left( 0 \\lt b \\lt \\dfrac{\\sqrt{2}}{2} \\text{\u306e\u3068\u304d} \\right) \\\\ a = \\dfrac{1}{2} & \\left( b \\geqq \\dfrac{\\sqrt{2}}{2} \\text{\u306e\u3068\u304d} \\right) \\end{array}\\right.}\r\n\\]\r\n

        (2)<\/strong><\/p>\r\n

        \\(b = \\dfrac{1}{\\sqrt{3}}\\) \u306e\u3068\u304d, (1)<\/strong> \u306e\u7d50\u679c\u3088\u308a\r\n\\[\r\na = \\dfrac{1}{\\sqrt{3}} \\cdot \\sqrt{1 -\\dfrac{1}{3}} = \\dfrac{\\sqrt{2}}{3}\r\n\\]\r\n\u306a\u306e\u3067, \u63a5\u70b9\u306e \\(x\\) \u5ea7\u6a19 \\(p\\) \u306f\r\n\\[\r\np = \\dfrac{\\frac{\\sqrt{2}}{3}}{1 -\\frac{1}{3}} = \\dfrac{\\sqrt{2}}{2}\r\n\\]\r\n\u63a5\u70b9\u306f, \\(C _ 2\\) \u4e0a\u306e\u70b9\u306a\u306e\u3067, \\(y\\) \u5ea7\u6a19 \\(q\\) \u306f\r\n\\[\r\nq = \\sqrt{\\dfrac{1 -\\frac{1}{2}}{3}} = \\dfrac{\\sqrt{6}}{6}\r\n\\]\r\n\u3088\u3063\u3066, \u63a5\u70b9\u306e\u5ea7\u6a19\u306f\r\n\\[\r\n\\underline{\\left( \\dfrac{\\sqrt{2}}{2} , \\dfrac{\\sqrt{6}}{6} \\right)}\r\n\\]\r\n

        (3)<\/strong><\/p>\r\n\"toko_2013_05_02\"\r\n

        \u9762\u7a4d\u3092\u6c42\u3081\u308b\u90e8\u5206\u306f \\(x\\) \u8ef8\u306b\u3064\u3044\u3066\u5bfe\u79f0\u306a\u306e\u3067, \\(y \\geqq 0\\) \u306e\u90e8\u5206\u306b\u3064\u3044\u3066\u8003\u3048\u308b.
        \r\n\u76f4\u7dda \\(x = \\dfrac{\\sqrt{2}}{2}\\) \u3068 \\(x\\) \u8ef8\u3068 \\(C _ 1\\) , \u76f4\u7dda \\(x = \\dfrac{\\sqrt{2}}{2}\\) \u3068 \\(x\\) \u8ef8\u3068 \\(C _ 2\\) \u306b\u56f2\u307e\u308c\u305f\u90e8\u5206\u306e\u9762\u7a4d\u3092\u305d\u308c\u305e\u308c \\(S _ 1 , S _ 2\\) \u3068\u304a\u304f\u3068, \u6c42\u3081\u308b\u9762\u7a4d \\(S\\) \u306f\r\n\\[\r\nS = 2 ( S _ 2 -S _ 1 ) \\quad ... [3]\r\n\\]\r\n\\(S _ 1\\) \u306b\u3064\u3044\u3066, \u3053\u306e\u90e8\u5206\u306f\u4e0b\u56f3\u659c\u7dda\u90e8\u306e\u3088\u3046\u306b\u306a\u308b\u306e\u3067<\/p>\r\n\"toko_2013_05_03\"\r\n

        \\[\\begin{align}\r\nS _ 1 & = \\dfrac{1}{2} \\left( \\dfrac{\\sqrt{2}}{3} \\right)^2 \\dfrac{\\pi}{3} -\\dfrac{1}{2} \\cdot \\dfrac{\\sqrt{2}}{6} \\cdot \\dfrac{\\sqrt{6}}{6} \\\\\r\n&= \\dfrac{\\pi}{27} -\\dfrac{\\sqrt{3}}{36}\r\n\\end{align}\\]\r\n\u307e\u305f\r\n\\[\\begin{align}\r\nS _ 2 & = \\displaystyle\\int _ {\\frac{\\sqrt{2}}{2}}^{1} \\sqrt{\\dfrac{1-x^2}{3}} \\, dx \\\\\r\n& = \\dfrac{\\sqrt{3}}{3} \\underline{\\displaystyle\\int _ {\\frac{\\sqrt{2}}{2}}^{1} \\sqrt{1-x^2} \\, dx} _ {[4]}\r\n\\end{align}\\]\r\n\u4e0b\u7dda\u90e8 [4] \u306f, \u4e0b\u56f3\u659c\u7dda\u90e8\u306e\u9762\u7a4d\u3092\u8868\u3059\u306e\u3067<\/p>\r\n\"toko_2013_05_04\"\r\n

        \\[\\begin{align}\r\n[4] & = \\dfrac{1}{2} \\cdot 1^2 \\cdot \\dfrac{\\pi}{4} -\\dfrac{1}{2} \\cdot \\dfrac{\\sqrt{2}}{2} \\cdot \\dfrac{\\sqrt{2}}{2} \\\\\r\n& = \\dfrac{\\pi}{8} -\\dfrac{1}{4}\r\n\\end{align}\\]\r\n\u3057\u305f\u304c\u3063\u3066\r\n\\[\r\nS _ 2 = \\dfrac{\\sqrt{3} \\pi}{24} -\\dfrac{\\sqrt{3}}{12}\r\n\\]\r\n\u3088\u3063\u3066, [3] \u3088\u308a, \u6c42\u3081\u308b\u9762\u7a4d\u306f\r\n\\[\\begin{align}\r\nS & = 2 \\left\\{ \\left( \\dfrac{\\sqrt{3} \\pi}{24} -\\dfrac{\\sqrt{3}}{12} \\right) -\\left( \\dfrac{\\pi}{27} -\\dfrac{\\sqrt{3}}{36} \\right) \\right\\} \\\\\r\n& = \\underline{\\dfrac{9 \\sqrt{3} -8}{108} \\pi -\\dfrac{\\sqrt{3}}{9}}\r\n\\end{align}\\]\r\n","protected":false},"excerpt":{"rendered":"\\(a , b\\) \u3092\u6b63\u306e\u5b9f\u6570\u3068\u3057, \u5186 \\(C _ 1 : \\ (x-a)^2+y^2 = a^2\\) \u3068\u6955\u5186 \\(C _ 2 : \\ x^2+\\dfrac{y^2}{b^2} = 1\\) \u3092\u8003\u3048\u308b. (1)\u3000\\(C … \u7d9a\u304d\u3092\u8aad\u3080 →<\/span><\/a>","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"inline_featured_image":false,"footnotes":""},"categories":[92],"tags":[141,111],"class_list":["post-806","post","type-post","status-publish","format-standard","hentry","category-toko_2013","tag-toko","tag-111"],"_links":{"self":[{"href":"https:\/\/www.roundown.net\/nyushi\/wp-json\/wp\/v2\/posts\/806","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.roundown.net\/nyushi\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.roundown.net\/nyushi\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.roundown.net\/nyushi\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.roundown.net\/nyushi\/wp-json\/wp\/v2\/comments?post=806"}],"version-history":[{"count":0,"href":"https:\/\/www.roundown.net\/nyushi\/wp-json\/wp\/v2\/posts\/806\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.roundown.net\/nyushi\/wp-json\/wp\/v2\/media?parent=806"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.roundown.net\/nyushi\/wp-json\/wp\/v2\/categories?post=806"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.roundown.net\/nyushi\/wp-json\/wp\/v2\/tags?post=806"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}