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大阪大学 前期理系 1992年度 問1
解答の著作権は解答作成者に帰属します。無断転載、複製を禁止します。
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 | 前期理系 |
| 年度 | 1992年度 |
| 問No | 問1 |
| 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 |    |
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\begin{FRAME}
$f(x),\,\,g(x)$ を2次関数とし，
2つの放物線 $F : y = f(x),\,\,\,G : y = g(x)$ を考える．
ただし，$F$ は下に凸で原点Oを頂点とし，
$G$ は上に凸でその頂点AはOと異なるものとする．
$G$ の上の点Pを直線OA上にはないようにとる．
点Oを通り直線APに平行な直線と $F$ との交点のうち，
O以外の点をQとする．
さらに，
直線OAと直線PQの交点をRとする．\smallskip
このとき，\smallskip
線分の長さの比 $\dfrac{\A\R}{\O\R}$ は点Pのとり方に関係なく
一定であることを示せ．
\end{FRAME}
\noindent{\color[named]{BurntOrange}\bfseries \fbox{解答}}
\vskip 2mm
$\A(\alpha,\,\,\beta)$とし，
\begin{align*}
f(x) = a x^2,\quad
g(x) = -b(x - \alpha)^2 + \beta,\quad
a > 0,\quad
b > 0
\end{align*}
とおく．
直線APの傾きを $m$ とする．\smallskip
$\dfrac{\A\R}{\O\R}$ が $m$ の値によらず一定値をとることを示せばよい．
このときAPの式は $y = m(x - \alpha) + \beta$ である．
Pの$x$座標は \\
\begin{minipage}{240pt}
\begin{align*}
-b(x - \alpha)^2 + \beta = m(x - \alpha) + \beta \\
\therefore \,\,\,
(x - \alpha)\{b(x - \alpha) + m\} = 0
\end{align*}
の $\alpha$ と異なる解だから，
\begin{gather*}
b(x - \alpha) + m = 0 \quad より \\[1mm]
x = -\frac{m}{b} + \alpha
\end{gather*}
$\O\Q \heikou \A\P$ だからOQの式は $y = mx$．\\
よってQの$x$座標は，
\begin{align*}
ax^2 = mx \qquad
\therefore \,\,\,
x(ax - m) = 0
\end{align*}
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\vskip 0.5zw
\noindent%
の0と異なる解だから，
$x = \dfrac{m}{a}$．\smallskip
$\triangle\O\Q\R\,\souzi\,\triangle\A\P\R$ より
\begin{align*}
\O\R : \A\R
&= \O\Q : \A\P \\
&= \sqrt{\vphantom{b} 1 + m^2}\,
\zettaiti{\dfrac{m}{a}}
: \sqrt{\vphantom{b} 1 + m^2}\,
\zettaiti{\left(-\dfrac{m}{b} + \alpha\right) - \alpha} \\[1mm]
&= \frac{m}{a} : \frac{m}{b} \\
&= b : a \\
\therefore \,\,\,
\frac{\A\R}{\O\R}
&= \frac{a}{b}
\end{align*}
これは $m$ の値によらず一定である．
\hfill ■
\end{document}
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