(1) 4x2−9y2=36x29−y24=14x^2 - 9y^2 = 36 \qquad \dfrac{x^2}{9} - \dfrac{y^2}{4} = 14x2−9y2=369x2−4y2=1 a=3a=3a=3、b=2b=2b=2 から漸近線は y=±23xy=\pm\dfrac{2}{3}xy=±32x (2) −4x2+9y2=36x2−9+y24=1-4x^2 + 9y^2 = 36 \qquad \dfrac{x^2}{-9} + \dfrac{y^2}{4} = 1−4x2+9y2=36−9x2+4y2=1 a=−3a=-3a=−3、b=2b=2b=2 から漸近線は y=±23xy=\pm\dfrac{2}{3}xy=±32x (3) x2−2y2=1x21−y212=1x^2 - 2y^2 = 1 \qquad \dfrac{x^2}{1} - \dfrac{y^2}{\frac{1}{2}} = 1x2−2y2=11x2−21y2=1 a=1a=1a=1、b=12b=\dfrac{1}{\sqrt{2}}b=21 から漸近線は y=±12xy=\pm\dfrac{1}{\sqrt{2}}xy=±21x (4) −x2+2y2=1x2−1+y212=1-x^2 + 2y^2 = 1 \qquad \dfrac{x^2}{-1} + \dfrac{y^2}{\frac{1}{2}} = 1−x2+2y2=1−1x2+21y2=1 a=−1a=-1a=−1、b=12b=\dfrac{1}{\sqrt{2}}b=21 から漸近線は y=±12xy=\pm\dfrac{1}{\sqrt{2}}xy=±21x グラフは問題集 p.142 を参照