Question

已知x=acosθ,y=bsinθ,求证：x^2/a^2+y^2/b^2=1

拜托了！谢谢！！！

Answer 1

x=acosθ， X^2 =a^2(cosθ)^2 , 等式两边除以 a^2 ,得 x^2/a^2 =(cosθ)^2 , --------(1)y=bsinθ , y^2=b^2 (sinθ)^2 , 等式两边除以 b^2 得 y^2/b^2 = (sinθ)^2 -------(2)(1)+(2) 得x^2/a^2+y^2/b^2==(cosθ)^2+(sinθ)^2=1

Answer 2

把代入等式的左边化简下答案就出来了