High school math, the conic curve part, the best calculation, don t give an idea, I calculate it mys

Solution: The ellipse intersects with the x-axis and y-axis at a(0,-1), respectively

The distance from the two points that l intersect the ellipse to a is equal.

The perpendicular bisector of the line segment formed by these two points crosses point A.

Simultaneous y = kx + b and x + 3y = 3

Then 3=x +3(kx+b) = (1+3k)x +6kbx+3b, i.e., (1+3k) x +6kbx+3(b -1)=0

Let the two intersections be (x1, y1) and (x2, y2).

then x1+x2=-6kb (1+3k), x1x2=3(b -1) (1+3k).

Then y1+y2=(kx1+b)+(kx2+b)=k(x1+x2)+2b=-6k b (1+3k)+2b=2b (1+3k).

There are 2 intersections.

Discriminant =36k b -12b -36k b +12+36k =36k -12b +12>0

3k²-b²+1>0

The coordinates of the midpoint of the two intersections are (-3kb (1+3k), b (1+3k)).

The vertical bisector is perpendicular to l, and the slope is -1 k

and the perpendicular bisector crosses point a, and the equation is y-(-1)=-1 k (x-0), i.e., y=-(1 k)x-1

Again, the midpoint is on the perpendicular bisector.

b/(1+3k²)=-(1/k)[-3kb/(1+3k²)]1

b=(3k²+1)/2

Substituting the conclusion of the discriminant formula 3k -b +1>0

3k -(3k +1) 4+1>0, i.e. 3k -2k -1<0, i.e. (k -1) (3k +1) <0

3k²+1>0

k²-1<0

i.e. -1

The answer upstairs is incomplete, it should be (-1,0) and (0,1) solution: let the straight line l:y=kx+m be the straight line that satisfies the condition, and then let p be the midpoint of mn, and if you want to meet the condition, you only need ap mn.

Composed. y＝kx+m

x^2/3+y^2=1 ===>(1+3k^2)x^2+6mkx+3m^2-3=0

Let m(x1,y1) n(x2,y2).

xp=x1+x2/2=-3mk/(1+3k^2) yp=kxp+m=m/(1+3k^2)

kap=3k^2-m+1/3mk

By ap mn==>3k 2-m+1 3mk=-1 k(k≠0), so m=-(3k 2+1) 2

by =36m2

k24（1+3k2

3m23）=9（1+3k2

1-k20, we get -1 k 1, and k ≠0

Therefore, when k (-1,0) (0,1), there is a straight line l that satisfies the condition

The one who answered was really domineering, I couldn't hit some of them, and I just thought he was done.

丨f f 丨=2 6

丨pf 丨-丨pf 丨丨=4

The above formula is squared on both sides.

pf₁|²pf₂|2丨pf 丨丨pf 丨=16 using the cosine theorem.

24=丨pf 丨 +丨pf 丨 -丨pf 丨丨pf 丨.

Subtract the two formulas.

8=|pf₁||pf₂|

It can be obtained from the area formula.

1 2丨f f 丨h=1 2 8sin60 h=4 3 2 6= 2

So yp= 2

Use the equal area method. The area formula b of the hyperbolic focal triangle divides the tan angle f1pf2 and then uses the ordinary triangle to find the area method 1 2 base multiplied by the height. The two equations are equal to be solved. There are positive and negative.

<> if there is no miscalculation, it is like this.

This question needs to be noted.

Definition of hyperbola; 2, cosine theorem; 3.Triangle to find the area formula.

Summary. Kiss! Please take a picture of the original question and send it to me or take a picture and send it to me after handwriting clearly, the expression of the question should be standardized, the description should be complete, and the teacher can see and understand it clearly, so as to better help you.

Kiss! Please take a picture of the original problem and send it to me or take a picture and send it to me after handwriting clearly, the expression of the question should be standardized, the description should be complete, and the teacher Pi Pei can see and understand it clearly, so as to better help you.

The second question of 21 questions, look at my solution** has a problem.

Take a picture of the question and send it to me.

x1 plus x2 x1x2 is right for your old guess, I hope you can take a look at me** unburied boy wrong bend Hanwang, thank you (*°3

I can't read what you're writing.

You take a picture of the original question and send it to me.

You can't write clearly, and you can't shoot clearly, I can't see a clue.

Library. The second half of Document 22—24 pages.

It is more convenient to set the points p, a, and b, and the direct midpoint formula of pa, pb, and ab is more troublesome to set the line, no matter how p(m, n), y=k(x-m)+n is not easy to find two k according to the two midpoints, and it is necessary to ......

I have, all the notes, the knowledge points are relatively small, because there is no mess, skills are the last word, all kinds of conclusions.

Solution: Let m(x,y),p(s,t).

Then: the vector pm=(x s,y t), the vector ma=( x, 1 y) because the ratio of the point m to the vector pa is 2:1

i.e.: vector pm=2 vector ma=( 2x, 2 2y)=(x s,y t) so 2x=x s

2－2y＝y－t

i.e.: s=3x, t=3y2

i.e.: p(3x,3y+2).

Because p is the point above the parabola y=2x 2+1.

So: 3y+2=2(3x) +1

i.e.: y=6x 1 3

The trajectory equation for the point m is: y=6x 1 3

(1) by the title, a(-2,0), b(2,0), m(t,sqrt(4-t 2) 2), n(t,-sqrt(4-t 2) 2).

Let C1 (x1,0), C2 (x2,0), and the garden C1 with a radius of |ac1|=|mc1|==>(x1+2)^2=(t-x1)^2+1-t^2/4==>x1=3t/8-3/4.

To the park C2, radius |bc2|=|mc2|==>(x2-2)^2=(t-x2)^2+1-t^2/4==>x2=3t/8+3/4.

Therefore |c1c2|=|x2-x1|=3/2.

2) The sum of the area of C1 and C2 s=pi*(|ac1|^2+|bc2|^2)=pi*[(x1+2)^2+(x2-2)^2]

pi*[(3t/8-3/4+2)^2+(3t/8+3/4-2)^2]=pi*(9t^2+100)/32.

Because -2

1) When the tangent is 135°, the area is the smallest, p( 2 2, 2 2) substitution into the ellipse: 1 a2+1 b2=2,e=c a= 2 2,c2 a2=1 2 c2=a2 2=a2-b2,b2=a2 huzhi 2 b2=c2,b=c 1 a2+2 a2=2,3 a2=2,a2=3 2,b2=3 4, x2 (3 2)+y2 (3 4)=1,(2)upper vertex (0, 3 2), right focal point ( 3 2,0), vertical is related to vertical. NF135°,AB45°,Let AB equation y=x+m a(x1,y1),b(x2,y2), kaf=y1 (x1-c),knb=(y2-b) x2,b=c y1 (x1-c) (y2-b) x2=-1 y1y2-by1=-x1x2+cx2 y1=x1+m y2=x2+m y1y2=x1x2+m(x1+x2)+m2 y1+y2=x1+x2+2m x2 a2+(x+m)2 b2=1 b2x2+a2 (take it as fiber x+m)2=a2b2 a2=2b2 substitution:

b2x2+2b2（x+m）2=2b 4 x2+2（x+m）2=2b2 3x2+4mx+2m2-2b2=0 x1+x2=-4m 3，x1x2=2（m2-b2） 3 y1y2=x1x2+m（x1+x2）+m2 =2（m2-b2） 3+m（-4m 3）+m2 =（2 3-4 3+1）m2-2b2 3 =（m2-2b2） 3 y1y2-by1=-x1x2+cx2 （m2-2b2） 3-by1=-2（m2-b2） 3+bx2 m2-4b2 3=b（y1+x2） =b（x1+x2+m） =b（-4m 3+m） m2-4b2 3=-4bm 3+bm m2+bm 3-4b2 3=0 3m2+bm-4b2=0 （3m+4b）（m-b）=0 m1=-4b 3=（-4 3）（ 3 2）=-2 3 m2=b= 3 2 m2=b， NAB co-point, imitation is not in line with the topic, m=-2 3 3 y=x-2 3 3

The relationship between the hyperbolic focal point and the real axis and the imaginary axis is: c 2 = a 2 + b 2, from which we can find the beam without bond coordinates of the left Jiaocha dry point a, and the coordinates of the f point are (-c, y), which is substituted into the hyperbolic equation to find the value of y. The value of y is the distance of af.