Two hyperbolic problems in senior high school mathematics, one expert can solve 200 points! ! ! Answer quickly and you can add more! ! !
(1) solution: branch 1, A (-a, 0), B (0, b), F (c, 0). = = >; 3ac=ba*bf=(-a,-b)*(c,-b)=b^2-ac.= = = = = & gt； b^2=4ac.= = = & gtc^2-a^2=4ac.= = = & gte^2-4e- 1=0.= = = & gtE=2+√5。 (2) Solution: If the parabolic equation is x2 = 2py (p > 0), then 2pm=9, (p/2)+m=5. (m & gt0).= = = & gtP= 1，m=9/2。 Or p=9, m= 1/2. If the parabolic equation is x 2 =-2py, (p > 0). Then 2pm =-9. (p/2)-m = 5。 (m； p=4，m= 2√6