Question

已知直线L与双曲线x^2/2-y^2/3=1相交,并且过右焦点f2,所交弦长为3√2,求直线L方程.

Answer 1

F(√5,0)

y=k(x-√5)

x^2/2-y^2/3=1

3x^2-2[k(x-√5)]^2=6

(3-2k^2)x^2+4√5k^2*x-6-10k^2=0

x1+x2=

x1*x2=

(x1-x2)^2=(x1+x2)^2-4x1*x2=

(x1-x2)^2+(y1-y2)^2=(1+k^2)*(x1-x2)^2=(3√2)^2

k=