设P,Q,R是等轴双曲线上的任意三点,求证,三角形PQR的垂心H必在同一等轴双曲线上.（用参数方程解）

Question

设P,Q,R是等轴双曲线上的任意三点,求证,三角形PQR的垂心H必在同一等轴双曲线上.（用参数方程解）

Answer 1

参数方程可设为x = a/cos(t),y = a·tan(t).
并设P,Q,R分别对应t = 2p,2q,2r.
PQ斜率为(tan(2q)-tan(2p))/(1/cos(2q)-1/cos(2p))
= (cos(2p)sin(2q)-cos(2q)sin(2p))/(cos(2p)-cos(2q))
= sin(2q-2p)/(cos(2p)-cos(2q))
= 2sin(q-p)cos(q-p)/(2sin(q+p)sin(q-p))
= cos(q-p)/sin(q+p).
设b = cos(p)cos(q)cos(r)+cos(p)sin(q)sin(r)+sin(p)cos(q)sin(r)+sin(p)sin(q)cos(r),
c = sin(p)sin(q)sin(r)+sin(p)cos(q)cos(r)+cos(p)sin(q)cos(r)+cos(p)cos(q)sin(r).
取s = arctan(-b/c),则b·cos(s)+c·sin(s) = 0.
该式可变形为:cos(q-p)cos(s-r)+sin(q+p)sin(s+r) = 0 ①,
cos(p-r)cos(s-q)+sin(p+r)sin(s+q) = 0 ②,
cos(r-q)cos(s-p)+sin(r+q)sin(s+p) = 0 ③.
取S为曲线上对应t = 2s的点,可知RS斜率为cos(s-r)/sin(s+r).
而由①可得RS ⊥ PQ,同理②保证QS ⊥ RP,而③保证PS ⊥ QR.
因此S就是△PQR的垂心H,故H也在曲线上.