2016年11月8日火曜日

数学 - 放物線・楕円・双曲線 - 2次曲線 – 放物線・楕円・双曲線(楕円、焦点、頂点、中心、長軸、短軸)

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

2. $\begin{array}{l}c=\sqrt{4-1}=\sqrt{3}\\ 頂点\left(±1,0\right),\left(0,±2\right)\\ 焦点\left(0,±\sqrt{3}\right)\end{array}$

3. $\begin{array}{l}\frac{{x}^{2}}{25}+\frac{{y}^{2}}{\frac{25}{4}}=1\\ c=\sqrt{25-\frac{25}{4}}=\frac{5\sqrt{3}}{2}\\ 頂点\left(±5,0\right),\left(0,±\frac{5}{2}\right)\\ 焦点\left(±\frac{5\sqrt{3}}{2},0\right)\end{array}$

4. $\begin{array}{l}c=\sqrt{5-4}=1\\ 頂点\left(±2,0\right),\left(0,±\sqrt{5}\right)\\ 焦点\left(0,±1\right)\end{array}$

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

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

3. $\begin{array}{l}c=\frac{4}{2}=2\\ a=\sqrt{2}b\\ {b}^{2}+4=2{b}^{2}\\ {b}^{2}=4\\ b=2\\ a=2\sqrt{2}\\ \frac{{x}^{2}}{8}+\frac{{y}^{2}}{4}=1\end{array}$

4. $\begin{array}{l}d=\sqrt{1+{\left(\sqrt{3}+2\right)}^{2}}+\sqrt{1+{\left(-\sqrt{3}+2\right)}^{2}}\\ =\sqrt{8+4\sqrt{3}}+\sqrt{8-4\sqrt{3}}\\ =c\\ \sqrt{{x}^{2}+{\left(y-\sqrt{3}\right)}^{2}}+\sqrt{{x}^{2}+{\left(y+\sqrt{3}\right)}^{2}}=d\\ \sqrt{{x}^{2}+{\left(y+\sqrt{3}\right)}^{2}}=d-\sqrt{{x}^{2}+{\left(y-\sqrt{3}\right)}^{2}}\\ {x}^{2}+{\left(y+\sqrt{3}\right)}^{2}={d}^{2}+\left({x}^{2}+{\left(y-\sqrt{3}\right)}^{2}\right)-2d\sqrt{{x}^{2}+{\left(y-\sqrt{3}\right)}^{2}}\\ 4\sqrt{3}y={d}^{2}-2d\sqrt{{x}^{2}+{\left(y-\sqrt{3}\right)}^{2}}\\ 2d\sqrt{{x}^{2}+{\left(y-\sqrt{3}\right)}^{2}}={d}^{2}-4\sqrt{3}y\\ 4{d}^{2}\left({x}^{2}+{\left(y-\sqrt{3}\right)}^{2}\right)={d}^{4}-8\sqrt{3}y{d}^{2}+48{y}^{2}\\ 4{d}^{2}{x}^{2}+\left(4{d}^{2}-48\right){y}^{2}={d}^{2}\left({d}^{2}-12\right)\\ {d}^{2}=4{\left(\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}\right)}^{2}\\ =4\left(4+2\sqrt{4-3}\right)\\ =24\\ \frac{4·24{x}^{2}}{24·12}+\frac{4·12{y}^{2}}{24·12}=1\\ \frac{{x}^{2}}{3}+\frac{{y}^{2}}{6}=1\end{array}$

$\begin{array}{l}P\left(u,0\right),Q\left(0,v\right),R\left(x,y\right)\\ {u}^{2}+{v}^{2}={6}^{2}\\ x=\frac{2u}{3},y=\frac{v}{3}\\ u=\frac{3x}{2},v=3y\\ \frac{9}{4}{x}^{2}+9{y}^{2}=36\\ \frac{{x}^{2}}{16}+\frac{{y}^{2}}{4}=1\end{array}$