Question

求以椭圆3x^2+12y^2=39的焦点为焦点,以直线y=±x/2为渐近线的双曲线方程

Answer 1

∵x^2/13+y^2/(13/4)=1.∴a^2-13,b^2=13/4,a>b,焦点在X轴上.

c2=a2-b^2=13-13/4=39/4.

c=±√39/2.

由渐近线 y=±x/2得：b/a=1/2.a=2b

双曲线的焦半径c,c^2==a ^2+b^2=39/4.

(2b)^2+b^2=39/4.

5b^2=39/4,

b^2=39/20.

a^2=(2b)^2=4b^2=39/5

∴所求双曲线方程为：x^2/(39/5-y^2/(39/20)=1.