x^2-2(4tanθ+3)x+25sec^2θ=0的两根在复平面上的对应点为F1F2,一椭圆以F1F2为焦点,且过原点,则椭圆长轴方程x^2-2(4tanθ+3)x+25（secθ）^2=0(0≤θ≤π/4)的两根在复平面上的对应点为F1F2,一椭圆以F1F2为焦点

Question

x^2-2(4tanθ+3)x+25sec^2θ=0的两根在复平面上的对应点为F1F2,一椭圆以F1F2为焦点,且过原点,则椭圆长轴
方程x^2-2(4tanθ+3)x+25（secθ）^2=0(0≤θ≤π/4)的两根在复平面上的对应点为F1F2,一椭圆以F1F2为焦点,且过原点,则椭圆长轴长为?

Answer 1

x^2-2(4tant+3)x+25sec^2t=0
x^2-2(4tant+3)x+25(1+tan^2t)=0
x=(2(4tant+3)±√(4(16tan^2t+24tant+9)-4*25(1+tan^2t)))/2
=(4tant+3)±√(-9tan^2t+24tant-16)
=4tant+3±(3tant-4)i
2a=|F1O|+|F2O|=2√(4tant+3)^2+(3tant-4)^2=2√(25tan^2t+25)=10√(tan^2t+1)=10sect