D﹙m? m? +m﹢? ﹚
When y=0, x = m (m+ 1)
X =- 1 or x=2m+ 1.
Therefore, a is a constant (﹙ 1, 0 ﹚, b ﹙ 2m+ 1, 0).
1)m=3/2,D(3/2,25/8)
tanADH = AH/DH =( 1+ 1.5)/(25/8)= 4/5
2) When ∠ ADB = 60, an equilateral triangle
√3﹙ 1+m﹚=? m? +m﹢?
M =-1+2 √ 3 or-1.
Because B is on the right of A, M =- 1+2 √ 3.
When ∠ ADB = 90, isosceles right triangle
1+m=? m? +m﹢?
Similarly, M= 1.
So 1 ≤ m ≤- 1+2 √ 3.
3)AB=2m+2,OC=m+? ,S 1=(2m+2)(m+? )/2,
DH intersects BC at F, CT is perpendicular to DH and DG is perpendicular to CB.
FH/CO=BH/BO
(m+ 1)/(2m+ 1)= FH/( 1/2+ 1)
DF=DH-FH=(m? +2m+ 1)/2-(m? + 1.5m+0.5)/(2m+ 1)=[(m+ 1)? (2m+ 1)-(2m? +3m+ 1)]⊙[2(2m+ 1)]
DF=(m+ 1)m/2
S2=S triangle DCF+S triangle DFB=BO×DF=[(2m+ 1)×(m? +m)/2]/2=S 1
m=2,- 1,- 1/2
So m= 1.
BO=3,CO= 1.5,BC=(3√5)/2
S2= 15/2
So DG=2√5,
So the distance is 2√5.
I made it myself. It should be right.