In the triangle PMN with area of 1, Tan ∠ PMN = 1 / 2, Tan ∠ PNM = - 2, a suitable coordinate system is established, and the hyperbolic equation with m, n as the focus and passing through point P is obtained

In the triangle PMN with area of 1, Tan ∠ PMN = 1 / 2, Tan ∠ PNM = - 2, a suitable coordinate system is established, and the hyperbolic equation with m, n as the focus and passing through point P is obtained


If the height of Mn side is h, Mn = H / (Tan ∠ PMN) + H / (Tan ∠ PNM) = 3H / 2 (this formula is best derived by drawing a graph. First, it is understood by acute angle triangle, and then extended to obtuse angle triangle). The area of triangle PMN is s = Mn * H / 2 = 3H ^ 2 / 4 = 1, and H = 2 / sqrt (3), Mn = sqrt (3). (sqrt = root sign) so the ellipse



Given the angle MPN, ad is on PM, C and B are on PN, a and B intersect CD on F. if pf bisects the angle MPN, the proof is: 1 / PA + 1 / Pb = 1 / PC + 1 / PD


The three sides of △ PAB are cut by the straight line CD at C, F and D. according to Menelaus theorem: (AD / PD) * (BC / PC) * (FB / FA) = 1. Because pf bisection ∠ APB is obtained by angular bisection theorem: FB / FA = Pb / PA, the above two formulas are: AD / PA * PD = BC / Pb * PC, that is, (pd-pa) / PA * PD = (pb-pc) / Pb * PC, so 1 / pa - 1 / PD



Point O is on the bisector of angle MPN, and circle O intersects PN and pm at points a, B and C, D respectively


Send me a picture