{"id":143,"date":"2011-12-01T14:28:51","date_gmt":"2011-12-01T05:28:51","guid":{"rendered":"http:\/\/roundown.main.jp\/nyushi\/?p=143"},"modified":"2021-10-20T15:52:39","modified_gmt":"2021-10-20T06:52:39","slug":"ykr201004","status":"publish","type":"post","link":"https:\/\/www.roundown.net\/nyushi\/ykr201004\/","title":{"rendered":"\u6a2a\u56fd\u5927\u7406\u7cfb2010\uff1a\u7b2c4\u554f"},"content":{"rendered":"
\n

\\(a , b\\) \u3092\u6b63\u306e\u5b9f\u6570\u3068\u3059\u308b.\r\n\u66f2\u7dda\r\n\\[\r\nC : \\ \\dfrac{x^2}{a^2} + \\dfrac{( y-b )^2}{b^2} = 1\r\n\\]\r\n\u306f\u9818\u57df \\(D : \\ x^2+y^2 \\leqq 1\\) \u306b\u542b\u307e\u308c\u3066\u3044\u308b. \u6b21\u306e\u554f\u3044\u306b\u7b54\u3048\u3088.<\/p>\r\n

    \r\n

  1. (1)<\/strong>\u3000\\(( a , b )\\) \u304c\u5b58\u5728\u3059\u308b\u7bc4\u56f2\u3092 \\(ab\\) \u5e73\u9762\u4e0a\u306b\u56f3\u793a\u305b\u3088.<\/p><\/li>\r\n

  2. (2)<\/strong>\u3000\\(C\\) \u304c\u56f2\u3080\u90e8\u5206\u306e\u9762\u7a4d\u304c\u6700\u5927\u3068\u306a\u308b\u3068\u304d\u306e \\(a , b\\) \u306e\u5024\u3092\u6c42\u3081\u3088.<\/p><\/li>\r\n<\/ol>\r\n

    \r\n\r\n

    \u3010 \u89e3 \u7b54 \u3011<\/h4>\r\n

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

    \\(C\\) \u4e0a\u306e\u70b9\u306f, \\(( x , y ) = \\left( a \\cos \\theta , b ( 1 +\\sin \\theta ) \\right)\\) \u3068\u8868\u305b\u308b.
    \r\n\\(D\\) \u306e\u5f0f\u306b\u4ee3\u5165\u3057\u3066\r\n\\[\\begin{align}\r\na^2 \\cos^2 \\theta +b^2 ( 1 +\\sin \\theta )^2 & \\leqq 1 \\\\\r\na^2 ( 1 -\\sin^2 \\theta ) +b^2 ( 1 +2\\sin \\theta +\\sin^2 \\theta) & \\leqq 1 \\\\\r\n\\text{\u2234} \\quad ( b^2 -a^2 ) \\sin^2 \\theta +2b^2 \\sin \\theta +a^2 +b^2 -1 & \\leqq 0\r\n\\end{align}\\]\r\n\\(t = \\sin \\theta\\) \u3068\u304a\u304f\u3068, \\(-1 \\leqq t \\leqq 1\\) \u3067\u3042\u308a\r\n\\[\r\n( b^2 -a^2 ) t^2 +2b^2 t +a^2 +b^2 -1 \\leqq 0 \\quad ... [\\text{A}]\r\n\\]\r\n\u3057\u305f\u304c\u3063\u3066, \\(-1 \\leqq t \\leqq 1\\) \u306b\u304a\u3044\u3066[A]\u304c\u5e38\u306b\u6210\u7acb\u3059\u308b\u305f\u3081\u306e\u6761\u4ef6\u3092\u6c42\u3081\u308c\u3070\u3088\u3044.
    \r\n[A] \u306e\u5de6\u8fba\u3092 \\(f(t)\\) \u3068\u304a\u304f.\r\n\\[\r\nf(-1) = -1 \\lt 0 , \\ f(1) = 4b^2 -1 = (2b+1)(2b-1)\r\n\\]\r\n

      \r\n

    1. 1*<\/strong>\u3000\\(b^2 -a^2 \\geqq 0\\) \u3059\u306a\u308f\u3061 \\(b \\geqq a\\) \u306e\u3068\u304d
      \r\n\\(f(t)\\) \u306f\u4e0b\u306b\u51f8\u306e \\(2\\) \u6b21\u95a2\u6570, \u307e\u305f\u306f \\(1\\) \u6b21\u95a2\u6570\u306a\u306e\u3067, \u6761\u4ef6\u306f\r\n\\[\\begin{align}\r\nf(1) & = (2b+1)(2b-1) \\leqq 0 \\\\\r\n\\text{\u2234} \\quad 0 & \\lt b \\leqq \\dfrac{1}{2}\r\n\\end{align}\\]<\/li>\r\n

    2. 2*<\/strong>\u3000\\(b^2 -a^2 \\lt 0\\) \u3059\u306a\u308f\u3061 \\(b \\lt a\\) \u306e\u3068\u304d
      \r\n\\(f(t)\\) \u306f\u4e0a\u306b\u51f8\u306e \\(2\\) \u6b21\u95a2\u6570\u3068\u306a\u308b.\r\n\\[\r\nf(t) = \\left( b^2 -a^2 \\right) \\left( t -\\dfrac{b^2}{a^2 -b^2} \\right)^2 -\\dfrac{b^4}{a^2 -b^2} +a^2 +b^2 -1\r\n\\]\r\n\u8ef8\u306e\u4f4d\u7f6e\u306b\u3064\u3044\u3066\u8003\u3048\u308b\u3068,\r\n\\[\\begin{align}\r\n0 & \\leqq \\left| \\dfrac{b^2}{a^2 -b^2} \\right| \\leqq 1 \\\\\r\nb^2 & \\leqq a^2 -b^2 \\\\\r\n\\text{\u2234} \\quad 0 & \\lt b \\leqq \\dfrac{\\sqrt{2}a}{2}\r\n\\end{align}\\]\r\n

        \r\n

      1. (\u30a2)<\/strong>\u3000\\(0 \\lt b \\leqq \\dfrac{\\sqrt{2}a}{2}\\) \u306e\u3068\u304d
        \r\n\u5224\u5225\u5f0f\u306b\u3064\u3044\u3066\r\n\\[\\begin{align}\r\nD\/4 & = b^4 -( b^2 -a^2 )( a^2 +b^2 -1 ) \\\\\r\n& = b^4 -( b^4 -a^4 +a^2 -b^2 ) \\\\\r\n& = a^4 -a^2 +b^2 \\leqq 0 \\\\\r\n\\text{\u2234} \\quad b^2 & \\leqq a^2 \\left( 1 -a^2 \\right) \\\\\r\n\\text{\u2234} \\quad b & \\leqq a \\sqrt{1 -a^2}\r\n\\end{align}\\]\r\n\u3053\u3053\u3067 \\(g(a) = a \\sqrt{1 -a^2}\\) \u3068\u304a\u304f.
        \r\n\u5b9a\u7fa9\u57df\u306f \\(0 \\lt a \\leqq 1\\) \u3067\u3042\u308a\r\n\\[\\begin{align}\r\ng'(a) & = 1 \\cdot \\sqrt{1 -a^2} + a \\cdot \\dfrac{(-2a)}{2 \\sqrt{1 -a^2}} \\\\\r\n& = \\dfrac{(1 -a^2) -a^2}{\\sqrt{1 -a^2}} = \\dfrac{1 -2 a^2}{\\sqrt{1 -a^2}}\r\n\\end{align}\\]\r\n\\(g'(a) = 0\\) \u3092\u89e3\u304f\u3068\r\n\\[\r\na = \\dfrac{\\sqrt{2}}{2}\r\n\\]\r\n\u3057\u305f\u304c\u3063\u3066, \u5897\u6e1b\u8868\u306f\u4e0b\u8868\u306e\u901a\u308a.\r\n\\[\r\n\\begin{array}{c|ccccc} t & 0 & \\cdots & \\dfrac{\\sqrt{2}}{2} & \\cdots & 1 \\\\ \\hline g'(a) & & + & 0 & - & \\\\ \\hline g(a) & (0) & \\nearrow & \\dfrac{1}{2} & \\searrow & 0 \\\\ \\end{array}\r\n\\]<\/li>\r\n

      2. (\u30a4)<\/strong>\u3000\\(\\dfrac{\\sqrt{2}a}{2} \\lt b \\lt a\\) \u306e\u3068\u304d\r\n\\[\\begin{align}\r\nf(1) & \\leqq 0 \\\\\r\n\\text{\u2234} \\quad 0 & \\lt b \\leqq \\dfrac{1}{2}\r\n\\end{align}\\]<\/li>\r\n<\/ol><\/p><\/li>\r\n<\/ol>\r\n

        \u4ee5\u4e0a\u3088\u308a, \u6c42\u3081\u308b\u9818\u57df\u306f\u4e0b\u56f3\u659c\u7dda\u90e8(\u305f\u3060\u3057, \u5b9f\u7dda\u5883\u754c\u306f\u542b\u307f, \u8ef8, \u25cb\u306f\u542b\u307e\u306a\u3044)<\/p>\r\n\"yokokoku2010_04_01\"\r\n

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

        \\(C\\) \u306e\u56f2\u3080\u90e8\u5206\u306e\u9762\u7a4d \\(S\\) \u306f\r\n\\[\r\nS = ab \\pi\r\n\\]\r\n\u3053\u308c\u3092\u5909\u5f62\u3057\u3066\r\n\\[\r\nb = \\dfrac{S}{\\pi a} \\quad ... [\\text{A}]\r\n\\]\r\n\u53cc\u66f2\u7dda P \u304c (1)<\/strong> \u3067\u6c42\u3081\u305f\u9818\u57df\u3068\u5171\u6709\u70b9\u3092\u3082\u3064\u3068\u304d\u306e\u3046\u3061, \\(S\\) \u304c\u6700\u5927\u3068\u306a\u308b\u3068\u304d\u3092\u8003\u3048\u308c\u3070\u3088\u3044.
        \r\n\u3053\u308c\u306f, \u9818\u57df\u306e\u5883\u754c\u7dda\u4e0a\u306b\u9650\u3089\u308c\u308b.<\/p>\r\n

          \r\n

        1. 1*<\/strong>\u3000\\(b = \\dfrac{1}{2}\\) \u306e\u3068\u304d
          \r\n\\(a\\) \u306e\u5909\u57df\u306f, \\(0 \\lt a \\leqq \\dfrac{\\sqrt{2}}{2}\\) \u306a\u306e\u3067\r\n\\[\r\nS = \\dfrac{\\pi a}{2} \\leqq \\dfrac{\\sqrt{2} \\pi}{4} \\quad \\left( \\text{\u7b49\u53f7\u6210\u7acb\u306f} \\ a = \\dfrac{\\sqrt{2}}{2} \\right)\r\n\\]<\/li>\r\n

        2. 2*<\/strong>\u3000\\(b = a \\sqrt{1 -a^2}\\) \u306e\u3068\u304d
          \r\n\\(a\\) \u306e\u5909\u57df\u306f, \\(\\dfrac{\\sqrt{2}}{2} \\leqq a \\lt 1\\) .\r\n\\[\r\nS = \\pi a^2 \\sqrt{1 -a^2}\r\n\\]\r\n\\(s = a^2\\) \u3068\u304a\u304f\u3068, \\(\\dfrac{1}{2} \\leqq p \\lt 1\\) \u3067\u3042\u308a\r\n\\[\r\n\\left( \\dfrac{S}{\\pi} \\right)^2 = a^4 -a^6 = s^2 -s^3\r\n\\]\r\n\u3053\u308c\u3092 \\(h(s)\\) \u3068\u304a\u3051\u3070\r\n\\[\r\nh'(s) = 2s -3s^2 = s ( 2 -3s )\r\n\\]\r\n\\(h'(s) = 0\\) \u3092\u89e3\u304f\u3068, \\(s = \\dfrac{2}{3}\\) .
          \r\n\u3057\u305f\u304c\u3063\u3066, \\(h(s)\\) \u306e\u5897\u6e1b\u8868\u306f\u4e0b\u8868\u306e\u901a\u308a\u3067\r\n\\[\r\n\\begin{array}{c|ccccc} s & \\dfrac{1}{2} & \\cdots & \\dfrac{2}{3} & \\cdots & 1 \\\\ \\hline h'(s) & & + & 0 & - & \\\\ \\hline h(s) & \\dfrac{1}{8} & \\nearrow & \\text{\u6700\u5927} & \\searrow & \\\\ \\end{array}\r\n\\]\r\n\u6700\u5927\u5024\u306f\r\n\\[\r\nh \\left( \\dfrac{2}{3} \\right) = \\dfrac{4}{27}\r\n\\]\r\n\u3057\u305f\u304c\u3063\u3066, \\(S\\) \u306f \\(a = \\sqrt{\\dfrac{2}{3}} = \\dfrac{\\sqrt{6}}{3}\\) \u306e\u3068\u304d\u306b\u6700\u5927\u5024\r\n\\[\r\n\\pi \\sqrt{\\dfrac{4}{27}} = \\dfrac{2\\sqrt{3} \\pi}{9}\r\n\\]\r\n\u3092\u3068\u308b.
          \r\n\u3053\u306e\u3068\u304d \\(b\\) \u306e\u5024\u306f\r\n\\[\r\nb = \\dfrac{\\sqrt{6}}{3} \\sqrt{1 -\\dfrac{2}{3}} = \\dfrac{\\sqrt{2}}{3}\r\n\\]<\/li>\r\n<\/ol>\r\n

          1*<\/strong> 2*<\/strong> \u3088\u308a, \u6c42\u3081\u308b\u5024\u306f\r\n\\[\r\n( a , b ) = \\underline{ \\left( \\dfrac{\\sqrt{6}}{3} , \\dfrac{\\sqrt{2}}{3} \\right)}\r\n\\]\r\n","protected":false},"excerpt":{"rendered":"\\(a , b\\) \u3092\u6b63\u306e\u5b9f\u6570\u3068\u3059\u308b. \u66f2\u7dda \\[ C : \\ \\dfrac{x^2}{a^2} + \\dfrac{( y-b )^2}{b^2} = 1 \\] \u306f\u9818\u57df \\(D : \\ x^2+y^2 \\leqq 1\\) … \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":[43],"tags":[14,9],"class_list":["post-143","post","type-post","status-publish","format-standard","hentry","category-yokokoku_r_2010","tag-14","tag-yokokoku_r"],"_links":{"self":[{"href":"https:\/\/www.roundown.net\/nyushi\/wp-json\/wp\/v2\/posts\/143","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=143"}],"version-history":[{"count":0,"href":"https:\/\/www.roundown.net\/nyushi\/wp-json\/wp\/v2\/posts\/143\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.roundown.net\/nyushi\/wp-json\/wp\/v2\/media?parent=143"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.roundown.net\/nyushi\/wp-json\/wp\/v2\/categories?post=143"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.roundown.net\/nyushi\/wp-json\/wp\/v2\/tags?post=143"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}