According to the second definition of ellipse, |AF|=e|AM|=ea, |BF|=e|BC|=eb.
| NE |/| AM | = | CE |/| CM | from△CEN∽△CMA？ ∵AM‖EF‖BC∴|CE|/|CM|=|BF|/|BA|
∴|NE|/|AM|=|BF|/|BA|？ | NE | = | AM | * | BF |/| BA | = a *(EB)/(ea+EB)= ab/(a+b)
Similarly | NF |/| BC | = | AF |/| AB | DE | NF | = | BC | * | AF |/| AB | = AB/(A+B)
∴|NE|=|NF|？ That is, the midpoint of AC divided by EF.
2. Solution: (1) The nature of the perpendicular from the middle |NP|=|PQ|? ∴|MP|+|NP|=|MP|+|PQ|=|MQ|=4√2
The locus of point p is an ellipse with m and n as the focus and 4√2 as the long axis.
The trajectory equation of point P is (y 2)/8+(x 2)/4 =1.
⑵ In △MQN, | Mn | 2 = | MQ | 2+| QN | 2-2 | MQ ||||| QN | COS ∠ Mqn is obtained by cosine theorem.
That is,16 = | MQ | 2 +| qn ||| 2-MQ |||| = (| MQ |+qn |) 2-3 | MQ ||||||| | qn | = (4 √ 2) 2-3 | MQ ||||||||
Solution | MQ ||| qn | =16/3
The area of the triangle MQN = 0.5 | MQ || qn | sin ∠ mqn = (4 ∠ 3)/3.
3. Solution: (1) b=c It can be known from "the quadrilateral formed by the short axis endpoint and the focus of an ellipse is a square"
∴a^2=b^2+c^2=2c^2
The distance between two lines of an ellipse is (2a 2)/c = (4c 2)/c = 4c = 4 ∴ c =1∴ a 2 = 2c 2 = 2? b=c= 1
The elliptic equation is x/2+y 2 =1?
(2) Direct narration is very troublesome, and it is easy to get dizzy when reading it. For convenience, I write the problem-solving process in the figure below.
Understand; Understanding