Question

设椭圆中心为原点O,一个焦点为F(0,1),长轴和短轴的长度之比为2：1 求椭圆方程

Answer 1

2a/2b=2:1

a=2b

c^2=a^2-b^2=3b^2

c=1,b^2=1/3 a^2=4/3

焦点在y轴

y^2/(4/3)+x^2/(1/3)=1

3y^2+12x^2=4

y=tx

3t^2x^2+12t^2x^2=4

x^2=4/(3t^2+12t^2)

y^2=4t^2/(3t^2+12t^2)

OQ^2=x^2+y^2=4(t^2+1)/(3t^2+12t^2)

OP^2=OQ^2*（2√3)^2=12*4(t^2+1)/(3t^2+12t^2)

Px^2+Py^2=(t^2+1)Px^2

Px^2=48/(3t^2+12t^2)

Px=4√3/√(3t^2+12t^2)

Py=4√3t/√(3t^2+12t^2)