Решебник по геометрии 11 класс. Атанасян ФГОС 866

\[\boxed{\mathbf{866.}ОК\ ГДЗ\ –\ домашка\ на\ 5}\]

\[Дано:\]

\[Гипербола;\]

\[\angle x,асим = 60{^\circ};\]

\[F_{1}F_{2} = 4.\]

\[Найти:\]

\[\textbf{а)}\ \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1;\]

\[\textbf{б)}\ e - ?;\]

\[\textbf{в)}\ x = \frac{a^{2}}{c}.\]

\[Решение.\]

\[1)\ Расстояние\ между\ фокусами\ \]

\[F_{1}F_{2} = 4:\]

\[c = OF_{1} = OF_{2} = 2.\]

\[2)\ tg\ \angle F_{2}OA = tg\ 60{^\circ} = \sqrt{3}\]

\[\frac{b}{a} = \sqrt{3}.\]

\[3)\ b^{2} = c^{2} - a^{2}\ \]

\[(по\ свойству\ гиперболы):\]

\[b^{2} = 4 - a^{2}\ \]

\[\frac{b^{2}}{a^{2}} = 3 = \frac{4 - a^{2}}{a^{2}}\]

\[3a^{2} = 4 - a^{2}\ \]

\[4a^{2} = 4\]

\[\ a^{2} = 1;\ \]

\[b^{2} = 4 - 1 = 3.\]

\[\textbf{а)}\ Каноническое\ уравнение\ \]

\[гиперболы:\]

\[\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1\ \]

\[x^{2} - \frac{y^{2}}{3} = 1.\]

\[\textbf{б)}\ Эксцентриситет\ равен:\ \ \]

\[e = \frac{c}{a} = \frac{2}{1} = 2.\]

\[\textbf{в)}\ Напишем\ уравнения\ \]

\[директрис:\]

\[x = \frac{a^{2}}{c}\]

\[x = - \frac{1}{2}\ \ и\ \ x = \frac{1}{2}.\]

\[\mathbf{Ответ}:\ \ а)\ x^{2} - \frac{y^{2}}{3} = 1;\ \ \]

\[\textbf{б)}\ e = 2;\ \ в)\ x = \pm \frac{1}{2}.\]