In the plane rectangular coordinate system xoy, it is known that the abscissa of a point m on the hyperbola x2 / 4-y2 / 12 = 1 is 3, then the distance from m to the right focus of the hyperbola is? If the abscissa of a point m on the hyperbola x2 / 4-y2 / 12 = 1 in the plane rectangular coordinate system xoy is 3, then the distance from m to the right focus of the hyperbola is? Through point a (60), make straight line L and hyperbola x2 / 16-y2 / 4 = 1 intersect at B C. two points a are the midpoint of line BC, then the equation of straight line L is? Through point a (6,1), make straight line L and hyperbola x2 / 16-y2 / 4 = 1 intersect at B C. two points a are the midpoint of line BC, then the equation of straight line L is?

1. Let the right focus coordinate f (C, 0), C ^ 2 = a ^ 2 + B ^ 2 = 4 + 12 = 16,
c=4,
Find the ordinate of M, 3 ^ 2 / 4-y ^ 2 / 12 = 1,
y=±√15,
If you want to find that the m-point is symmetrical up and down with the x-axis as the symmetry axis,
The right focus coordinate is (4,0),
|MF|=√（4-3)^2+(√15)^2=4,
The distance from m to the right focus of hyperbola is 4
2. Let B (x1, Y1), C (X2, Y2),
x1^2/16-y1^2/4=1,(1)
x2^2/16-y2^2/4=1,(2)
(1) (2) formula,
(x1^2-x2^2)/16-(y1^2-y2^2)/4=0,
1/4-[(y1-y2)/(x1-x2)]*[(y1+y2)/2]/[(x1+x2)/2]=0,（3）
Let the slope of the linear equation l be K, then k = (y1-y2) / (x1-x2),
Point a is the midpoint of BC, and the coordinates are: (x1 + x2) / 2 = 6,
(Y1 + Y2) / 2 = 1, (is it 0 or 1?)
Substituting (3),
1/4-k*1/6=0,
k=3/2,
Then the linear equation is: (Y-1) / (X-6) = 3 / 2,
y=3x/2-8.

RELATED INFORMATIONS