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)