大阪大学 後期理系 2002年度 問3

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解答作成者: 森 宏征

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入試情報

大学名 大阪大学
学科・方式 後期理系
年度 2002年度
問No 問3
学部 理学部 ・ 医学部 ・ 歯学部 ・ 薬学部 ・ 工学部 ・ 基礎工学部
カテゴリ 図形と方程式
状態 解答 解説なし ウォッチリスト

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\documentclass[a4paper,12pt,fleqn]{jreport} \setlength{\topmargin}{-25mm} \setlength{\oddsidemargin}{2.5mm} \setlength{\textwidth}{420pt} \setlength{\textheight}{700pt} \usepackage{amsmath} \usepackage{amssymb} \usepackage{ascmac} \usepackage{graphicx} \usepackage{delarray} \usepackage{multicol} \usepackage{amscd} \usepackage{pifont} \usepackage{color} \ExecuteOptions{usename} \usepackage{vector3} \usepackage{fancybox} \usepackage{custom_mori} \begin{document} \setlength{\abovedisplayskip}{0.5zw} \setlength{\belowdisplayskip}{0.5zw} \begin{FRAME}  $0 < t < 1$ とし,\smallskip 曲線 $y = \dfrac{1}{x}$ 上に点$\A\!\left(t,\,\,\dfrac{1}{t} \right)$をとる. Aを通り直線 $y = x$ に直交する直線と $y = x$ の交点をPとし, 原点に関してPと対称な点をQとする. \begin{enumerate} \item[(1)]  QとAを通る直線を $l$ とするとき, $l$ の傾きを求めよ. \item[(2)]  $l$ 上に点RをAに関してQの反対側にとる. $\angle\P\A\R$の二等分線が点Aにおけるこの曲線の接線と 直交するときの $t^2$ の値を求めよ. \end{enumerate} \end{FRAME} \noindent{\color[named]{BurntOrange}\bfseries \Ovalbox{解答}} \begin{enumerate} \renewcommand{\labelenumi}{(\arabic{enumi})} \item  Aを通り $y = x$ に直交する直線は \\ \begin{minipage}{230pt} \begin{gather*} y = -(x - t) + \frac{1}{t} \\[1mm] \therefore \,\,\, y = -x + t + \frac{1}{t} \end{gather*} よってPの$x$座標は \begin{gather*} -x + t + \frac{1}{t} = x \\[1mm] \therefore \,\,\, x = \frac{1}{2}\!\left(t + \frac{1}{t} \right) \end{gather*} ゆえに$\P\!\left(\dfrac{1}{2}\!\left(t+\dfrac{1}{t}\right),\,\, \dfrac{1}{2}\!\left(t+\dfrac{1}{t} \right) \right)$である. \end{minipage} \begin{minipage}{220pt} \hspace*{0zw} %\input{osaka02s3s_zu_3} %WinTpicVersion4.23 \unitlength 0.1in \begin{picture}( 20.1600, 21.1600)( 4.0000,-22.8400) % STR 2 0 3 0 Black White % 4 1372 191 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{\scriptsize$\theta$} \put(16.1000,-5.8000){\makebox(0,0)[lb]{{\scriptsize$\theta$}}}% % STR 2 0 3 0 Black White % 4 1640 744 1640 780 2 0 0 0 % {\scriptsize$\theta$} \put(16.4000,-7.8000){\makebox(0,0)[lb]{{\scriptsize$\theta$}}}% % STR 2 0 3 0 Black White % 4 1304 1164 1304 1200 2 0 0 0 % {\scriptsize$\pphi$} \put(13.0400,-12.0000){\makebox(0,0)[lb]{{\scriptsize$\pphi$}}}% % CIRCLE 3 0 3 0 Black White % 4 1336 475 1372 723 1314 835 1543 817 % {\color[named]{Black}{% \special{pn 4}% \special{ar 1336 476 252 252 1.0265158 1.6318315}% }}% % STR 2 0 3 0 Black White % 4 1430 1374 1430 1410 2 0 0 0 % {\footnotesize O} \put(14.3000,-14.1000){\makebox(0,0)[lb]{{\footnotesize O}}}% \end{picture}% \end{minipage} \vskip 0.5zw \noindent% Qは原点Oに関するPの対称点だから $\Q\!\left(-\dfrac{1}{2}\!\left(t+\dfrac{1}{t} \right),\,\, -\dfrac{1}{2}\!\left(t+\dfrac{1}{t} \right) \right)$. これより, \begin{align*} (lの傾き) = \frac{\,\dfrac{1}{\mathstrut t} + \dfrac{1}{\mathstrut 2}\!\left(t+\dfrac{1}{t} \right)\,} {\,t + \dfrac{\mathstrut 1}{2}\!\left(t+\dfrac{1}{t} \right)\,} = \frac{\,\dfrac{t}{\mathstrut 2} + \dfrac{3}{\mathstrut 2t}\,} {\,\dfrac{\mathstrut 3t}{2} + \dfrac{\mathstrut 1}{2t}\,} = \color{red}{\boldsymbol{\frac{t^2 + 3}{3t^2 + 1}}} \tag*{$\Ans$} \end{align*} \item  $\angle\P\A\R = 2\theta$, 直線APが$x$軸の正の向きとなす角を$-45^\circ$, $l$ が$x$軸となす角を $\pphi$ とすれば, \begin{gather*} 2\theta = \pphi - (-45^\circ) = \pphi + 45^\circ,\quad \tan\pphi = (lの傾き) = \frac{t^2 + 3}{3t^2 + 1} \end{gather*} よって, \begin{align*} \tan 2\theta = \tan(\pphi + 45^\circ) = \frac{\tan\pphi + 1}{1 - \tan\pphi} = \frac{\,\dfrac{t^2 + 3}{\mathstrut 3t^2 + 1} + 1\,} {\,1 - \dfrac{\mathstrut t^2 + 3}{3t^2 + 1}\,} = \frac{2t^2 + 2}{t^2 - 1} \end{align*} 一方で,$\tan 2\theta = \dfrac{2\tan\theta}{1 - \tan^2\theta}$ だから, \begin{gather*} \frac{2t^2 + 2}{t^2 - 1} = \frac{2\tan\theta}{1 - \tan^2\theta} \\ \therefore \,\,\, (t^2 - 1)\tan\theta = (t^2 + 1)(1 - \tan^2\theta) \tag*{$\cdott\MARU{1}$} \end{gather*} Aにおける $y = \dfrac{1}{x}$ の接線の傾きは $-\dfrac{1}{t^2}$. $\angle\P\A\R$の二等分線の傾きは \begin{align*} \tan(\theta - 45^\circ) = \frac{\tan\theta - 1}{1 + \tan\theta} \end{align*} $\angle\P\A\R$の二等分線がAでの接線と直交するから, \begin{gather*} -\frac{1}{t^2} \cdot \frac{\tan\theta - 1}{1 + \tan\theta} = -1 \qquad \tan\theta - 1 = t^2(\tan\theta + 1) \\[1mm] \therefore \,\,\, \tan\theta = -\frac{t^2+1}{t^2-1} \tag*{$\cdott\MARU{2}$} \end{gather*} \MARU{2}を\MARU{1}に代入して, \begin{gather*} (t^2 - 1) \cdot \left(-\frac{t^2 + 1}{t^2 - 1}\right) = (t^2 + 1) \cdot \frac{-4t^2}{(t^2 - 1)^2} \\ \qquad (t^2 - 1)^2 = 4t^2 \qquad t^4 - 6t^2 + 1 = 0 \\ \therefore \,\,\, t^2 = {\color{red}{\boldsymbol{3 - 2\sqrt{\vphantom{b} 2}}}} \quad (\,\because \,\,\,0 < t^2 < 1) \tag*{$\Ans$} \end{gather*} \end{enumerate} \end{document}