The second question:
( 1)
It can be seen from the question that (a+√3b)(a-√3b)=0.
Namely: (x+√3, √3y)(x-√3, -√3y)=0.
De: x2-3-3y2=0
So the trajectory equation of point P(x, y) is
x2/3-y2= 1
(2) 1. When the straight line L intersects with D(0,-1)
- 1=k*0+m
At this time m=- 1.
2. When the straight line L intersects the hyperbola.
By x2-3-3y2=0
Y=kx+m, you can get (1/3-k2) x2-2kmx-m2-1= 0 by eliminating y.
Let A(x, y), B(X, y), AB midpoint c.
xX =(-3 m2-3)/( 1-3 k2)÷0……①
△= 4k2m 2+4(m2+ 1)( 1/3-k2)÷0……②
X+x = 6km /( 1-3k2)
So the midpoint C{3km/( 1-3k2), m/( 1-3k2)}
So the perpendicular equation of AB is y-m/(1-3k2) =-1/k {x-3km/(1-3k2)}.
Because it passed D(0,-1)
Then:-1-m/(1-3k2) = 3km/(1-3k2), that is: 4m/(1-3k2) =-1... ③.
It can be obtained from ① ② ③: m > 4.
To sum up, the value range of m is {- 1} ∨( 4, +∞).