x^2/ 13-y^2/ 12=- 1

a^2= 12,b^2= 13

c^2=a^2+b^2= 12+ 13=25,c=5

Lower alignment equation: y =-a 2/c =- 12/5.

e=c/a=5/√ 13

Distance from point to focus =e* Distance from point to directrix

So e (y1+12/5)+e (y3+12/5) = 2e (6+12/5).

y 1+y3= 12

2)

x3^2/ 13-y3^2/ 12=- 1,x 1^2/ 13-y 1^2/ 13=- 1

(x3^2-x 1^2)/ 13=(y3^2-y 1^2)/ 12

Ac slope = (y3-y1)/(x3-x1) =12 (x1+x3)/13 (y1+y3) = (.

Ac midpoint coordinates ((x 1+x3)/2, (y 1+y3)/2) are: ((x1+x3)/2,6).

Slope of AC vertical line:-13/(x 1+x3)

Equation of the vertical line of AC: y =-13/(x1+x3) * (x-(x1+x3)/2)+6.

=- 13x/(x 1+x3)+ 13/2+6

=- 13x/(x 1+x3)+25/2

Namely: y+13x/(x1+x3) = 25/2.

Therefore, when x = 0 and y = 25/2, the equation holds.

That is, the middle vertical line of line segment AC passes through a fixed point.

(0,25/2)