求两条渐近线为x±2y=0且截直线x-y-3=0所得弦长为(8*√3)/3的双曲线方程尽快啦注:√3指根号3

Question

求两条渐近线为x±2y=0且截直线x-y-3=0所得弦长为(8*√3)/3的双曲线方程
尽快啦
注:√3指根号3

Answer 1

双曲线方程:x^2/a^2-y^2/b^2=1 ==>(bx)^2-(ay)^2=(ab)^2 .(1)
y=(1/2)x ==>b/a=1/2 ==>2b=a ==>4b^2=a^2 .(2)
(2)代入(1):
x^2-4y^2=4b^2 .(3)
x-y-3=0 ==>x=y+3 .(4)
(4)代入(3):
(y+3)^2-4y^2=4b^2 ===>3y^2-6y+4b^2-9=0 ===>y1+y2=2 y1y2=(4b^2-9)/3
(y1-y2)^2=(y1+y2)^2-4y1y2=4-4(4b^2-9)/3=16(3-b^2)/3
x1-x2=(y1+3)-(y2+3)=y1-y2 ==>(x1-x2)^2=(y1-y2)^2=16(3-b^2)/3
(x1-x2)^2+(y1-y2)^2=32(3-b^2)/3=(8√3/3)^2=64/3 ==>3-b^2=2 ==>b^2=1
a^2=4b^2=4
双曲线方程:x^2/4-y^2=1