From the topic P 1F2⊥F 1F2 vector P 1F 1* vector P 1F2=9/4, The area of △f 1 F2 is equal to 3/2 | p1f1|| p1F2 | cos θ = 9/4 | p1f1|| p/kloc. 2pf 1 = 5/2, pf2 = 3/2, H 1, H2 = 2, 2a=4, 2c=2, b=√3 The ellipse is: x 2/4+y 2/3 = 65438. MH = (qh1+ph2)/2 = pq/2e = apq/2c = | pq | Linear elliptic equation, q (x 1, y 1), p (x2, y2) (3+4k 2) x 2. x 1*x2=(4k^2- 12)/(3+4k^2 |x 1-x2|= 12sqrt(k^2+ 1)/(3+4k^2 |mh|= 1/sqrt(k^2+ 1 *|x 1-x2|= 12/(3+4k^2)①③。 OF 1=(- 1，0)，OQ=(x 1，y 1)，OG=(0，k) GQ/QF 1=λ，GQ/gf 1 =λ/( 1+λ)OQ = 1/( 1+λ)*(0，K)+λ/( 1+λ)*(-666