Vector PF 1 vector PF2=0

∴(x-c,y) (x+c，y)=0

∴x？ +y？ =c？

e 1=c/a 1，e2=c/a2

∴ 1/(e 1)？ + 1/(e2)？ =[(a 1)？ +(a2)？ ]/c？

All of the above can be deduced from the topic, so let's start to solve it (a 1)? +(a2)？ Use c? Relationship between

Let the elliptic equation be x? /(a 1)？ +y？ /(b 1)？ = 1; The equation of hyperbola is x? /(a2)？ -Really? /(b2)？ = 1

There is also a known uselessness: p is the intersection point.

Substitute p(x, y) into elliptic equation and hyperbolic equation.

Ellipse: (a 1)? -(b 1)？ =c？

The equation can be arranged as [(a 1)? -c？ ]x？ +(a 1)？ y？ =(a 1)？ [(a 1)？ -c？ ]

∵x？ +y？ =c？

∴ The equation is arranged as c? x？ =2(a 1)？ c？ -(a 1)？ (a 1)？ I can't play the fourth power. Hehe, forgive me. )

Similarly, the provable hyperbolic equation can be arranged as follows:

c？ x？ =2(a2)？ c？ -(a2)？ (a2)？

∴ Subtract two curve equations to get

(a 1)？ +(a2)？ =2c？

∴ 1/(e 1)？ + 1/(e2)？ =[(a 1)？ +(a2)？ ]/c？ =2

This kind of problem is mostly a large amount of calculation, not very skilled, so calm down and work it out.

2. extend the extension line of the intersection point DA and h of CE.

∴△AEH∽△BEC

∴AH:BC=AE:BE= 1/2

∴ vector AH= 1/2 vector BC= 1/2b.

You can also prove that △FGH∽△BGC.

FG:BG=FH:BC=3/4

∴FG=3/4BG=3/7FB=3/7 (vector a b+ vector FA)=3a/7-3b/28.

Vector AG= vector af+ vector fg = b/4 = 3a/7-3b/28 = 3a/7+b/7.

This method of doing the problem is generally cumbersome, that is, constantly inverting the vectors around the target vector. It is suggested to consider adding auxiliary lines when it cannot be reversed. Also, the draft of this question should be written neatly (what is a neat draft? ! ), otherwise sometimes you will find that the answer is upside down but the draft is full of spots. . . Hmm. How interesting

I am a freshman now. Looking at these senior three questions, I can't help but sigh: "Senior three questions are getting more and more abnormal! ! ! ! "