(2) A = (5 5,3) and k=xy= 15 after the change.

(3)y= 15/x

Points P (x 1, y 1), y 1=3, x 1=5, OD=t, BP = 5-t.

s 1 = 1/2×BD×BP = 1/2×3×(5-t)= 15-3t/2

The q point (x2, y2), x2=OD+CD=t+4, CQ = y2 =15/x2 =15/(t+4).

S2 = 1/2×CD×CQ = 1/2×4× 15/(t+4)= 30/(t+4)

(4)S2= 10/7S 1

30/(t+4)= 10/7[ 15-3t/2]

To simplify: t? -t-6=0

(t-3)(t+2)=0

T=3 or t=-2 (s)

When t=3, S2= 10/7S 1.