2c = the square root of 6 of 2

c=√6

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

Therefore, the elliptic equation is:

x^2/9+y^2/3= 1

(2) substitute y=kx-2 into the elliptic equation.

x^2+3(k^2x^2-4kx+4)-9=0

(3k^2+ 1)x^2- 12kx+3=0①

Let A(x 1, kx 1-2) and B(x2, kx2-2).

I Vieta theorem

x 1+x2= 12k/(3k^2+ 1)②

x 1x2=3/(3k^2+ 1)

︱PA︱=︱PB︱ Germany

(x 1-0)^2+(kx 1-2- 1)^2=(x2-0)^2+(kx2-2- 1)^2

(x 1-x2)*[(k^2+ 1)(x 1+x2)-6k]=0

Because the straight line y=kx-2 does not include the straight line x=0, x 1≠x2, so

(k^2+ 1)(x 1+x2)-6k=0

12k(k^2+ 1)/(3k^2+ 1)=6k

The solution is k 2 =1or k=0.

When k=0, equation ① becomes x 2+3 = 0. There is no solution. So k = 1

The equation of the straight line l is:

Y=x-2 or y=-x-2.