= & gt|BD|-|AD|=|AC|-|BC|=2

The equation of =>g is: y 2-x 2 =1(y

(2) Let a straight line be y = kx-2.

Equation brought into g:

(kx-2)^2 - x^2 = 1

(k^2- 1)*x^2 -4kx +3 =0

Let the roots be x 1, x2 and x 1=-3*x2 (from PA=3AQ).

x 1 + x2 = 4k/(k^2- 1)

x 1*x2 = 3/(k^2- 1)

It can be solved by the above three formulas: 5 * k 2 = 1.

k^2= 1/5

= & gtK= plus or minus 5 /5.

Just bring k into the linear equation.