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)

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数学読本〈3〉平面上のベクトル/複素数と放物線・楕円・双曲線/空間図形/2次曲線/数列 (松坂和夫(著)、岩波書店)の第12章(放物線・楕円・双曲線 - 2次曲線)、12.1(放物線・楕円・双曲線)、楕円、問4、5、6.を取り組んでみる。

問4.

$\begin{array}{l} c = \sqrt{9 - 4} = \sqrt{5} \\ 頂点 (\pm 3, 0), (0, \pm 2) \\ 焦点 (\pm \sqrt{5}, 0) \end{array}$
$\begin{array}{l} c = \sqrt{4 - 1} = \sqrt{3} \\ 頂点 (\pm 1, 0), (0, \pm 2) \\ 焦点 (0, \pm \sqrt{3}) \end{array}$
$\begin{array}{l} \frac{x^{2}}{25} + \frac{y^{2}}{\frac{25}{4}} = 1 \\ c = \sqrt{25 - \frac{25}{4}} = \frac{5 \sqrt{3}}{2} \\ 頂点 (\pm 5, 0), (0, \pm \frac{5}{2}) \\ 焦点 (\pm \frac{5 \sqrt{3}}{2}, 0) \end{array}$
$\begin{array}{l} c = \sqrt{5 - 4} = 1 \\ 頂点 (\pm 2, 0), (0, \pm \sqrt{5}) \\ 焦点 (0, \pm 1) \end{array}$

問5.

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

問6.

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

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