## 2016年11月13日日曜日

### 数学 - 放物線・楕円・双曲線 - 2次曲線 – 放物線・楕円・双曲線(双曲線、漸近線、頂点、焦点、直角双曲線)

1. $\begin{array}{l}漸近線\\ y=±\frac{3}{4}x\\ 頂点\\ \left(-4,0\right),\left(4,0\right)\\ 焦点\\ 3=\sqrt{{c}^{2}-16}\\ 9={c}^{2}-16\\ c=±5\\ \left(-5,0\right),\left(5,0\right)\end{array}$

2. $\begin{array}{l}漸近線\\ \frac{{x}^{2}}{4}-\frac{{y}^{2}}{9}=1\\ y=±\frac{3}{2}x\\ 頂点\\ \left(-2,0\right),\left(2,0\right)\\ 焦点\\ 9={c}^{2}-4\\ c=±\sqrt{13}\\ \left(-\sqrt{13},0\right),\left(\sqrt{13},0\right)\end{array}$

3. $\begin{array}{l}漸近線\\ y=±x\\ 頂点\\ \left(-1,0\right),\left(1,0\right)\\ 焦点\\ {c}^{2}=1+1=2\\ c=±\sqrt{2}\\ \left(-\sqrt{2},0\right),\left(\sqrt{2},0\right)\end{array}$

4. $\begin{array}{l}漸近線\\ \frac{{x}^{2}}{25}-\frac{{y}^{2}}{4}=1\\ y=±\frac{2}{5}x\\ 頂点\\ \left(-5,0\right),\left(5,0\right)\\ 焦点\\ {c}^{2}=29\\ c=±\sqrt{29}\\ \left(-\sqrt{29},0\right),\left(\sqrt{29},0\right)\end{array}$

5. $\begin{array}{l}漸近線\\ y=±\frac{\sqrt{6}}{\sqrt{3}}x=±\sqrt{2}x\\ 頂点\\ \left(0,\sqrt{6}\right),\left(0,-\sqrt{6}\right)\\ 焦点\\ c=±3\\ \left(0,3\right),\left(0,-3\right)\end{array}$

6. $\begin{array}{l}漸近線\\ \frac{{x}^{2}}{\frac{1}{4}}-\frac{{y}^{2}}{\frac{1}{9}}=-1\\ y=±\frac{\sqrt{\frac{1}{9}}}{\sqrt{\frac{1}{4}}}x=±\frac{2}{3}\\ 頂点\\ \left(0,\frac{1}{3}\right),\left(0,-\frac{1}{3}\right)\\ 焦点\\ c=±\sqrt{\frac{1}{4}+\frac{1}{9}}=±\sqrt{\frac{13}{36}}=±\frac{\sqrt{13}}{6}\\ \left(0,\frac{\sqrt{13}}{6}\right),\left(0,-\frac{\sqrt{13}}{6}\right)\end{array}$

1. $\begin{array}{l}b=2,c=3,\\ {a}^{2}=9-4=5\\ \frac{{x}^{2}}{5}-\frac{{y}^{2}}{4}=-1\end{array}$

2. $\begin{array}{l}a=1,\\ \frac{b}{a}=2,b=2\\ {x}^{2}-\frac{{y}^{2}}{4}=1\end{array}$

3. $\begin{array}{l}c=1\\ \frac{b}{a}=\frac{1}{2}\\ 2b=a\\ {a}^{2}+{b}^{2}=1\\ 4{b}^{2}+{b}^{2}=1\\ {b}^{2}=\frac{1}{5}\\ {a}^{2}=4{b}^{2}=\frac{4}{5}\\ \frac{5}{4}{x}^{2}-5{y}^{2}=1\\ 5{x}^{2}-20{y}^{2}=4\end{array}$

4. $\frac{{x}^{2}}{4}-\frac{{y}^{2}}{4}=1$

5. $\begin{array}{l}3=2a\\ a=\frac{3}{2}\\ \frac{2}{9}{x}^{2}-\frac{2}{9}{y}^{2}=-1\\ 2{x}^{2}-2{y}^{2}=-1\end{array}$

6. $\begin{array}{l}d=\sqrt{{3}^{2}+{4}^{2}}-\sqrt{{3}^{2}}=5-3=2\\ \sqrt{{x}^{2}+{\left(y-2\right)}^{2}}-\sqrt{{x}^{2}+{\left(y+2\right)}^{2}}=±2\\ \sqrt{{x}^{2}+{\left(y-2\right)}^{2}}=±2+\sqrt{{x}^{2}+{\left(y+2\right)}^{2}}\\ {x}^{2}+{\left(y-2\right)}^{2}=4±4\sqrt{{x}^{2}+{\left(y+2\right)}^{2}}+\left({x}^{2}+{\left(y+2\right)}^{2}\right)\\ -8y-4=±4\sqrt{{x}^{2}+{\left(y+2\right)}^{2}}\\ 2y+1=±\sqrt{{x}^{2}+{\left(y+2\right)}^{2}}\\ 4{y}^{2}+4y+1={x}^{2}+{\left(y+2\right)}^{2}\\ {x}^{2}-3{y}^{2}=-3\\ \frac{{x}^{2}}{3}-{y}^{2}=-1\end{array}$

$\begin{array}{l}A\left(-c,0\right),B\left(c,0\right),M\left(0,0\right),P\left(x,y\right)\\ P{M}^{2}={x}^{2}+{y}^{2}\\ PA=\sqrt{{\left(x+c\right)}^{2}+{y}^{2}}\\ PB=\sqrt{{\left(x-c\right)}^{2}+{y}^{2}}\\ PA·PB=\sqrt{{\left(x+c\right)}^{2}{\left(x-c\right)}^{2}+{y}^{2}\left({\left(x+c\right)}^{2}+{\left(x-c\right)}^{2}\right)+{y}^{4}}\\ P{M}^{4}={\left(PA·PB\right)}^{2}\\ {x}^{4}+2{x}^{2}{y}^{2}+{y}^{4}={\left({x}^{2}-{c}^{2}\right)}^{2}+{y}^{2}\left(2{x}^{2}+2{c}^{2}\right)+{y}^{4}\\ 2{x}^{2}{y}^{2}=-2{c}^{2}{x}^{2}+{c}^{4}+2{x}^{2}{y}^{2}+2{c}^{2}{y}^{2}\\ {c}^{2}{x}^{2}-{c}^{2}{y}^{2}={c}^{4}\\ \frac{{x}^{2}}{{c}^{2}}-\frac{{y}^{2}}{{c}^{2}}=1\end{array}$