According to the definition of ellipse, m+n = 2a (1).
In the triangle F 1PF2, m 2+n 2-2mncost = 4c 2...(2) t = angle F 1PF2.
( 1)^2-(2):2mn( 1+cost)=4(a^2-c^2)
-& gt； mn=2b^2/( 1+cost)
-& gt； 1/2mnsint=b^2*sint/( 1+cost)
-& gt； S=b^2*tan(t/2).！ ! !
Note that you must understand it to remember it.
If it is a hyperbola, then the triangle area is B 2 * COT.
It can also be deduced in the same way, mainly the application of cosine theorem.
If you want to know more, please see.
/s？ ie = GB 23 12 & amp； bs = % CD % D6 % D4 % B2 % D6 % D0j % BD % B9 % B5 % E3 % C8 % FD % BD % C7 % D0 % CE % B5 % C4 % C3 % E6 % BB % FD & amp； sr = & ampz = & ampcl = 3 & ampf = 8 & ampD6 % D4 % B2 % D6 % D0 % BD % B9 % B5 % E3 % C8 % FD % BD % C7 % D0 % CE % B5 % C4 % C3 % E6 % BB % FD & amp； ct=0