Let the focus of hyperbola be on the X axis, passing through points A( 1, 0) and B(- 1, 0), and P is any point on hyperbola different from AB. If the vertical center H of triangle APB is always on this hyperbola, find the standard equation of hyperbola.

Solution: Let P(x0, y0).

Because H is the heart, PH⊥AB, AH⊥BP.

AB is on the X axis, so PH is perpendicular to the X axis. There are only two intersections between the straight line where PH is located and the hyperbola, one is P(x0, y0) and the other is (x0, -y0) (symmetry of hyperbola). Because H is also on the hyperbola, the coordinate of H is (x0, -y0).

So the vector AH=(x0- 1, -y0) (minus the coordinates of point H and point A),

Vector BP=(x0+ 1, y0),

Because the vector AH⊥BP, ah BP = 0 (the inner product is 0),

Namely, (x0-1) * (x0+1)-y0 2 = 0,

Simplified to x0 2-y0 2 =1.

P is any point on the hyperbola. Any point on the hyperbola satisfies the above equation, that is, the equation of hyperbola is: x 2-y 2 = 1.