怎么求椭圆的焦点坐标?
来源:百度知道 编辑:UC知道 时间:2024/05/25 20:49:41
如何用A,B,C,D,E,F这些参数得到椭圆中心点的坐标,椭圆的长半轴,短半轴,和长轴与X轴的夹角?
谁来告诉我啊,说算法也行!
AX^2 + BXY + CY^2 + DX + EY + F [A不等于0,不妨设A>0]
= A{X^2 + BXY/A + [BY/(2A)]^2} - B^2Y^2/(4A) + CY^2 + DX + EY + F
= A{[X + BY/(2A)]^2 + D[X + BY/(2A)]/A} - DBY/(2A) + Y^2[C - B^2/(4A)] + EY + F
= A{[X + BY/(2A)]^2 + D[X + BY/(2A)]/A + [D/(2A)]^2} - D^2/(4A) + Y^2[C - B^2/(4A)] + Y[E - BD/(2A)] + F
= A{X + BY/(2A) + D/(2A)}^2 + Y^2[C - B^2/(4A)] + Y[E - BD/(2A)] + F - D^2/(4A) 【C - B^2/(4A)不等于0,因A>0,所以 C - B^2/(4A)>0】
= A{X + BY/(2A) + D/(2A)}^2 + [C - B^2/(4A)]{Y^2 + Y[E - BD/(2A)]/[C - B^2/(4A)]} + F - D^2/(4A)
= A{X + BY/(2A) + D/(2A)}^2 + [C - B^2/(4A)]{Y^2 + Y[E - BD/(2A)]/[C - B^2/(4A)] + {[E - BD/(2A)]/[2C - B^2/(2A)]}^2 } - [E - BD/(2A)]^2/[4C - B^2/A] + F - D^2/(4A)
= A{X + BY/(2A) + D/(2A)}^2 + [C - B^2/(4A)]{Y + [E - BD/(2A)]/[2C - B^2/(2A)]}^2 - [E - BD/(2A)]^2/[4C - B^2/A] + F - D^2/(4A)
[因A>0,所以,{[E - BD/(2A)]^2/[4C - B^2/A] - F + D^2/(4A)} > 0]
椭圆中心