Question

已知α属于[0,π/2],试讨论当α的值变化时,方程x^2×sinα+y^2×cosα=1表示曲线的形状.

Answer 1

α属于[0,π/2],则sinα和cosα都大等于0小等于1;

α=0时,sinα=0,cosα=1,方程为y^2=1,则y=±1,为平行于x轴的两条直线;

α∈(0,π/4)时,sinα＜cosα,为焦点在x轴上的椭圆;

α=π/4时,sinα=cosα=√2/2,为中心在原点,半径为2^(1/4)的圆;

α∈(π/4,π/2)时,sinα＞cosα,为焦点在y轴上的椭圆;

α=π/2时,sinα=1,cosα=0,方程为x^2=1,则x=±1,为平行于y轴的两条直线;