(1)<\/strong>\u3000\\(D\\) \u306e\u65b9\u7a0b\u5f0f\u3092\u6c42\u3081\u3088.<\/p><\/li>\r\n (2)<\/strong>\u3000\\(C , D\\) , \\(y\\) \u8ef8\u304a\u3088\u3073\u76f4\u7dda \\(x = \\dfrac{1}{2}\\) \u3067\u56f2\u307e\u308c\u305f\u90e8\u5206\u3092 \\(x\\) \u8ef8\u306e\u307e\u308f\u308a\u306b \\(1\\) \u56de\u8ee2\u3055\u305b\u3066\u3067\u304d\u308b\u7acb\u4f53\u306e\u4f53\u7a4d\u3092\u6c42\u3081\u3088.<\/p><\/li>\r\n<\/ol>\r\n (1)<\/strong><\/p>\r\n P , Q \u306e \\(x\\) \u5ea7\u6a19\u3092 \\(p , q \\ ( p \\gt q )\\) \u3068\u304a\u304f. (2)<\/strong><\/p>\r\n \u6c42\u3081\u308b\u4f53\u7a4d\u3092 \\(V\\) \u3068\u304a\u304f\u3068\r\n\\[\\begin{align}\r\nV & = \\pi \\displaystyle\\int _ 0^{\\frac{1}{2}} \\left\\{ \\left( x^2 +\\dfrac{1}{1 +4x^2} \\right)^2 -\\left( x^2 \\right)^2 \\right\\} \\, dx \\\\\r\n& = \\pi \\displaystyle\\int _ 0^{\\frac{1}{2}} \\left\\{ \\dfrac{1}{2} \\left( 1 -\\dfrac{1}{1 +4x^2} \\right) +\\dfrac{1}{( 1 +4x^2 )^2} \\right\\} \\, dx \\\\\r\n& = \\dfrac{\\pi}{4} -\\dfrac{\\pi}{2} \\underline{\\displaystyle\\int _ 0^{\\frac{1}{2}} \\dfrac{1}{1 +4x^2} \\, dx} _ {[5]} +\\pi \\underline{\\displaystyle\\int _ 0^{\\frac{1}{2}} \\dfrac{1}{(1 +4x^2)^2} \\, dx} _ {[6]}\r\n\\end{align}\\]\r\n\u3053\u3053\u3067, [5] [6] \u306b\u3064\u3044\u3066, \\(x = \\dfrac{\\tan \\theta}{2} \\ \\left( -\\dfrac{\\pi}{2} \\lt \\theta \\lt \\dfrac{\\pi}{2} \\right)\\) \u3068\u304a\u304f\u3068\r\n\\[\r\ndx = \\dfrac{d \\theta}{2 \\cos^2 \\theta} , \\quad \\begin{array}{c|ccc} x & 0 & \\rightarrow & \\dfrac{1}{2} \\\\ \\hline \\theta & 0 & \\rightarrow & \\dfrac{\\pi}{4} \\end{array}\r\n\\]\r\n\u306a\u306e\u3067\r\n\\[\\begin{align}\r\n[5] & = \\dfrac{1}{2} \\displaystyle\\int _ 0^{\\frac{\\pi}{4}} \\dfrac{1}{1 +\\tan^2 \\theta} \\cdot \\dfrac{1}{\\cos^2 \\theta} \\, d \\theta \\\\\r\n& = \\dfrac{1}{2} \\left[ \\theta \\right] _ 0^{\\frac{\\pi}{4}} = \\dfrac{\\pi}{8} , \\\\\r\n[6] & = \\dfrac{1}{2} \\displaystyle\\int _ 0^{\\frac{\\pi}{4}} \\dfrac{1}{\\left( 1 +\\tan^2 \\theta \\right)^2} \\cdot \\dfrac{1}{\\cos^2 \\theta} \\, d \\theta \\\\\r\n& = \\dfrac{1}{2} \\displaystyle\\int _ 0^{\\frac{\\pi}{4}} \\cos^2 \\theta \\, d \\theta \\\\\r\n& = \\dfrac{1}{4} \\displaystyle\\int _ 0^{\\frac{\\pi}{4}} ( 1 +\\cos 2 \\theta ) \\, d \\theta \\\\\r\n& = \\dfrac{1}{4} \\left[ \\theta +\\dfrac{\\sin 2 \\theta}{2} \\right] _ 0^{\\frac{\\pi}{4}} \\\\\r\n& = \\dfrac{1}{4} \\left( \\dfrac{\\pi}{4} +\\dfrac{1}{2} \\right) = \\dfrac{\\pi}{16} +\\dfrac{1}{8}\r\n\\end{align}\\]\r\n\u3088\u3063\u3066\r\n\\[\\begin{align}\r\nV & = \\dfrac{\\pi}{4} -\\dfrac{\\pi}{2} \\cdot \\dfrac{\\pi}{8} +\\pi \\left( \\dfrac{\\pi}{16} +\\dfrac{1}{8} \\right) \\\\\r\n& = \\underline{\\dfrac{3 \\pi}{8}}\r\n\\end{align}\\]\r\n","protected":false},"excerpt":{"rendered":"\\(xy\\) \u5e73\u9762\u4e0a\u306b\u66f2\u7dda \\(C : \\ y = x^2\\) \u304c\u3042\u308b. \\(C\\) \u4e0a\u306e \\(2\\) \u70b9 P , Q \u304c \\(\\text{PQ} = 2\\) \u3092\u307f\u305f\u3057\u306a\u304c\u3089\u52d5\u304f\u3068\u304d, PQ \u306e\u4e2d\u70b9\u306e\u8ecc\u8de1\u3092 \\(D\\) … \u7d9a\u304d\u3092\u8aad\u3080
\r\n\r\n\u3010 \u89e3 \u7b54 \u3011<\/h4>\r\n
\r\n\u3055\u3089\u306b, \\(s = p+q\\) , \\(t = pq\\) \u3068\u304a\u304f\u3068, \\(p , q\\) \u306f\u65b9\u7a0b\u5f0f \\(u^ 2 -su +t = 0\\) \u306e\u7570\u306a\u308b \\(2\\) \u3064\u306e\u5b9f\u6570\u89e3\u3067\u3042\u308b\u306e\u3067, \u5224\u5225\u5f0f \\(D\\) \u306b\u3064\u3044\u3066\r\n\\[\r\nD = s^2 -4t \\gt 0 \\quad ... [1]\r\n\\]\r\n\u3092\u307f\u305f\u3059.
\r\nPQ \u306e\u4e2d\u70b9M \\(( X , Y )\\) \u306f\r\n\\[\\begin{align}\r\nX & = \\dfrac{p+q}{2} = \\dfrac{s}{2} , \\\\\r\nY & = \\dfrac{p^2 +q^2}{2} = \\dfrac{s^2 -2t}{2}\r\n\\end{align}\\]\r\n\u306a\u306e\u3067\r\n\\[\\begin{align}\r\ns & = 2X \\quad ... [2] , \\\\\r\nt & = \\dfrac{(2X)^2}{2} -Y =2X^2 -Y \\quad ... [3]\r\n\\end{align}\\]\r\n[1] \u306b [2] [3] \u3092\u4ee3\u5165\u3059\u308c\u3070\r\n\\[\\begin{align}\r\n4X^2 -4 ( 2X^2 -Y ) & \\gt 0 \\\\\r\n-X^2 +Y & \\gt 0 \\\\\r\n\\text{\u2234} \\quad Y & \\gt X^2 \\quad ... [4]\r\n\\end{align}\\]\r\n\\(\\text{PQ} = 2\\) \u3088\u308a\r\n\\[\\begin{align}\r\n( p-q )^2 +( p^2 +q^2 )^2 & = 4 \\\\\r\n( p-q )^2 \\left\\{ 1 +( p+q )^2 \\right\\} & = 4 \\\\\r\n\\text{\u2234} \\quad ( s^2 -4t ) ( 1 +s^2 ) & = 4\r\n\\end{align}\\]\r\n\u3053\u308c\u306b, [2] [3] \u3092\u4ee3\u5165\u3059\u308c\u3070\r\n\\[\\begin{align}\r\n\\left\\{ 4 X^2 -4 ( 2X^2 -Y ) \\right\\} ( 1 +4X^2 ) & = 4 \\\\\r\n4 ( Y^2 -X^2 ) & = \\dfrac{4}{1 +4X^2} \\\\\r\n\\text{\u2234} \\quad Y = X^2 & +\\dfrac{4}{1 +4X^2}\r\n\\end{align}\\]\r\n\u3053\u308c\u306f [4] \u3092\u307f\u305f\u3057\u3066\u3044\u308b.
\r\n\u3088\u3063\u3066, \u6c42\u3081\u308b \\(D\\) \u306e\u5f0f\u306f\r\n\\[\r\n\\underline{y = x^2 +\\dfrac{1}{1 +4x^2}}\r\n\\]\r\n![\"ykr20140501\"](\"\/\/www.roundown.net\/nyushi\/wp-content\/uploads\/ykr20140501.svg\")
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