数学 - 2次曲線と直線 - 2次曲線 – 2次曲線と直線(放物線と直線、接戦)

数学読本〈3〉

学習環境

• Surface 3 (4G LTE)、Surface 3 タイプ カバー、Surface ペン(端末)
• Windows 10 Pro (OS)
• 数式入力ソフト(TeX, MathML): MathType
• MathML対応ブラウザ: Firefox、Safari
• MathML非対応ブラウザ(Microsoft Edge(対応?), Google Chrome...)用JavaScript Library: MathJax

数学読本〈3〉平面上のベクトル/複素数と2次曲線と直線/空間図形/2次曲線/数列 (松坂 和夫(著)、岩波書店)の第12章(2次曲線と直線 - 2次曲線)、12.2(2次曲線と直線)、放物線と直線、問11、12、13、14.を取り組んでみる。

問11.

1. 
   $\begin{array}{l}4x^2+k^2+4kx=-8x\\ 4x^2+2\left(2k+4\right)x+k^2=0\\ \frac{D}{4}=4{\left(k+2\right)}^2-4k^2\\ =4\left(4k+4\right)\\ k>-1\\ 2\\ k=-1\\ 1\\ k<-1\\ 0\end{array}$
2. 
   $\begin{array}{l}{m}^2{\left(x-1\right)}^2=-8x\\ m^2{x}^2+\left(-2m^2+8\right)x+m^2=0\\ \frac{D}{4}={\left(m^2-4\right)}^2-m^4\\ =-8m^2+16\\ =-8\left(m^2-2\right)\\ 0<\left|m\right|<\sqrt{2}\\ 2\\ \left|m\right|=0,\sqrt{2}\\ 1\\ \left|m\right|>\sqrt{2}\\ 0\end{array}$

問12

$\begin{array}{l}y-1=a\left(x+1\right)\\ y=a\left(x+1\right)+1\\ a^2{\left(x+1\right)}^2+1+2a\left(x+1\right)=8x\\ a^2{x}^2+\left(2a^2+2a-8\right)x+a^2+1+2a=0\\ \frac{D}{4}={\left(a^2+a-4\right)}^2-a^2\left(a^2+2a+1\right)\\ =-8a^2-8a+16\\ =-8\left(a^2+a-2\right)\\ =-8\left(a+2\right)\left(a-1\right)\\ a=-2,1\\ y=-2x-1\\ y=x+2\end{array}$

問13.

$\begin{array}{l}P\left(-p,b\right)\\ y-b=m\left(x+p\right)\\ y=m\left(x+p\right)+b\\ m^2{\left(x+p\right)}^2+b^2+2bm\left(x+p\right)=4px\\ m^2{x}^2+\left(2p{m}^2+2bm-4p\right)x+m^2{p}^2+b^2+2bmp=0\\ \frac{D}{4}={\left(p{m}^2+bm-2p\right)}^2-m^2\left(m^2{p}^2+b^2+2bmp\right)\\ =-4p^2{m}^2-4bpm+4p^2\\ =-4p\left(p{m}^2+bm-p\right)\\ p{m}^2+bm-p=0\\ m_1{m}_2=-1\end{array}$

問14.

$\begin{array}{l}M\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)\\ y=\frac{y_1+y_2}{2}\\ y_{1}y=2\left(x+x_1\right)\\ y_{2}y=2\left(x+x_2\right)\\ \left(y_1-y_2\right)y=2\left(x_1-x_2\right)\\ y=\frac{2\left(x_1-x_2\right)}{y_1-y_2}\\ =\frac{2\left(\frac{y_1{}^2}{4}-\frac{y_2{}^2}{4}\right)}{y_1-y_2}\\ =\frac{\frac{1}{2}\left(y_1{}^2-y_2{}^2\right)}{y_1-y_2}\\ =\frac{y_1+y_2}{2}\end{array}$