The ultimate BOSS of difficult mathematics! Wait online! Dear God, help! ! ! (high school ellipse)

& lt 1 & gt； a(a^2/c,0)； B(0，b)； F(c，0)

Now find the coordinate (c, y) of point P: C2/A2+Y2/B2 =1.

So y = b 2/a

Point D is FP, and the midpoint coordinate is (c, B4/2a2); D is on AB, FD/OB = AF/AO.

(b 2/2a)/b = (b 2/c)/(a 2/c) So b/a= 1/2 So e=c/a = √3/2.

& lt2>A 1(-a, 0) A2(a, 0), B(0, b) vector BA 1(-a,-b); Vector BA2(a, -b)

Vector BA 1* vector ba2 =-a 2+b 2 =-3 from the first question a=2b.

So-4b 2+b 2 =-3 3; b= 1，a=2

Elliptic equation x 2/4+y 2 =1.

& lt3 & gtM(x 1，y 1)； There is an answer to the question N(x2, y2) on the Internet, but the calculation of the standard answer to the third question is more complicated, and the author will give a simple solution later.

Let the straight line A 1M: my = (x+2) be substituted into x 2+4y 2 = 4.

(my-2)^2+4y^2=4； y 1 =(m^2+4)/4m； x 1 = my 1-2 = (m^2-4)/4

Let the straight line A2N: ny = x-2 be substituted into X 2+4Y 2 = 4.

(ny+2)^2+4y^2=4； y2 = -(n^2+4)/4m； x2 =ny2+2 = (4-n^2)/4

And a1m. A2N crosses q; The alignment equation is x=4/√3.

So there is yq = (4/√ 3+2)/m = (4/√ 3-2)/n.

So m = (4/√ 3+2) k; n = (4/√3-2)k

Line Mn: (y-y1) * (x1-x2)/(y1-y2) = (x-x1) and then start tedious substitution, which is almost impossible, although the standard answer is to simplify it to X = 2 (M. _ & lt

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Parameter form M( 2cosα, sinα); N( 2cosβ, sinβ) requires proficiency and differential product formula.

Row a1m: y = (x+2) * [sinα/(2cosα+2)] = 2sinα/2cosα/2 * (x+2)/[4 (cos α/2) 2]

= (x+2)*(tanα/2)/2

Similarly A2N: y = (x-2)*/(-tanβ/2)*2.

Because A 1M, A2N crosses point q; so(4/√3+2)tanα/2 =- 1/tanβ/2(4/√3-2)。

tanα/2*tanβ/2 = -7+4√3

Now start looking for the straight line MN:

kMN =(sinα-sinβ)/(2 cosα-2 cosβ)= 2 sin(α-β)/2 * cos(α+β)/2/[-4 sin(α-β)/2 * sin(α+β)/2]

= -(cot(α+β)/2)/2

Line Mn: y-sinα = (x-2cosα) kmn

When y=0, x = 2 (sin α sin (α+β)/2)/cos (α+β)/2+2cosα.