(12) It is known that the center of hyperbola E is the origin, f (3,0) is the focus, the straight line L passing through f intersects hyperbola E at point A and point B, and the midpoint of AB is N(- 12,-15), then the equation of hyperbola E is
Solution:
According to the known conditions, the intersection point F is on the X axis, and the equation of hyperbola can be set as
x^2/a^2-y^2/b^2= 1，
Because c=3, therefore, A 2+B 2 = C 2 = 9, ...
The coordinates of point A and point B are A(x 1, y 1) and B(x2, y2) respectively.
From the midpoint of A and B is N(- 12,-15):
x 1+x2 = 2 *( 12)=-24，
y 1+y2 = 2 *( 15)=-30，…②
Let the slope of the straight line L be k, because both F and N are on the straight line L, so
k =(0+ 15)/(3+ 12)= 1，…③
A and b are both on hyperbola, so
x 1^2/a^2-y 1^2/b^2= 1，
x2^2/a^2-y2^2/b^2= 1，
Subtract the two expressions: (x12-x2 2)/a2-(y12-y2 2)/B2 = 0,
(x 1^2-x2^2)/a^2=(y 1^2-y2^2)/b^2，
(x 1-x2)(x 1+x2)/a^2=(y 1-y2)(y 1+y2)/b^2，
(x 1+x2)/(y 1+y2)*b^2/a^2=(y 1-y2)/(x 1-x2)，
Because A and B are on the straight line L, (y1-y2)/(x1-x2) = k =1,(combined with ③ formula).
Therefore, (x1+x2)/(y1+y2) * B2/a2 =1,
So B2/A2 = (y1+y2)/(x1+x2),
Combined with formula (2), b 2/a 2 = 5/4,
Combined with the formula (1), a 2 = 4, b 2 = 5,
So the hyperbolic equation is: x 2/4-y 2/5 = 1.