The ellipse 16x 2 25y 2 400 has ABC on it, where A 4, 12 5, ABC center of gravity is in the left foc

The BC equation is 20x+15y+68=0

KBC=-4 3, the coordinates of the midpoint of BC are (-5, 2, -6, 5), let B(X1, Y1), C(X2, Y2), and the left focus of the ellipse be (-3, 0).

The bc coordinates are substituted into the elliptic equation 16x1 2+25y1 2=400, and 16x2 2+25y2 2=400 are subtracted to obtain (y2-y1) (x2-x1)=-4 3

i.e. the slope is -4 3

The coordinates of the center of gravity are the arithmetic mean of the vertex coordinates, i.e., their coordinates are ((x1+x2-4) 3,(y1+y2+12 5) 3).

So x1+x2-4) 3=-3, y1+y2+12 5) 3=0, i.e. x1+x2=-5, y1+y2=-12 5

So the coordinates of the midpoint of BC are (-5 2, -6 5).

Did I make it clear?

The center of gravity coordinates g((-4+xb+xc) 3,(12 5+yb+yc) 3) is (-3,0), and the coordinates of the midpoint of bc (-5 2, -6 5) are obtained, and the coordinates of the two points of bc are set and substituted into the elliptic equation, and the gradient of bc is obtained k=(yb-yc) (xb-xc)=-16 25*(xb+xc) (yb+yc)=-16 25*(-5)*(12 5)=-4 3, so the linear bc equation is y=-4 3(x+5 2)-6 5 is y=-4 3x-68 15

25x^2+16y^2=1

A 2 = 1 16, B 2 = 1 25 1 16>1 25 The focus is on the Y axis of the tour.

c^2=1/16-1/25=9/16*25c=3/20

The focus coordinates are: (0,3 20)Sell search (0,-3 20).

The elliptic c equation x 2 4 + y 2 3 = 1, the left and right endpoints are a(-2,0), b(2,0).

c = (a 2-b 2) = 4-3) = 1 with the right focus of f(1,0).

Let the straight line through the right focal point be y=k(x-1).

Substituting the elliptic equation gives x 2 4+k 2(x-1) 2 3=1

Sorted out to 3x 2+4k 2(x-1) 2-12=0

3+4k^2)x^2-8k^2x+4k^2-12=0 (1)

Solving this one-dimensional quadratic equation yields two solutions of x, x1 and x2;

At the same time, there are x1+x2=8k 2 (3+4k 2), x1x2=(4k 2-12) (3+4k 2).

Substituting y=k(x-1) can solve two solutions of y, y1=k(x1-1), y2=k(x2-1).

Let the intersection of the line and the ellipse be m(x1,y1),n(x2,y2).

Then the equation for the straight line am is: y=y1 (x1+2)*(x+2).

The equation for the straight line bn is: y=y2 (x2-2)*(x-2).

The intersection p can be solved from two straight lines: y1 (x1+2)*(x+2)=y2 (x2-2)*(x-2).

y1/y2=(x1+2)/(x2-2)*(x-2)/(x+2)=(x1-1)/(x2-1)

x-2)/(x+2)=(x1-1)/(x2-1)*(x2-2)/(x1+2)

1-4/(x+2)=1-(x1+3x2-4)/(x1x2-x1+2x2-2)

x+2)/4=(x1x2-x1+2x2-2)/(x1+3x2-4)

x=4(x1x2-x1+2x2-2)/(x1+3x2-4)-2

x=4[2x1x2-3x1+x2]/(2x1+6x2-8)

2x1x2-3x1+x2)-(2x1+6x2-8)

2x1x2-5(x1+x2)+8

2*4(k^2-3)/(3+4k^2)-5*8k^2/(3+4k^2)+8

8[5k^2-k^2+3]/(3+4k^2)+8

2x1x2-3x1+x2)=(2x1+6x2-8)

2x1x2-3x1+x2)/(2x1+6x2-8)=1

x=4 substituting the solution yields y=6y1 (x1+2)=2y2 (x2-2).

That is, the abscissa of the intersection point is x=4, and the ordinate changes with the change of the point m,n, that is, the straight line x=4

The intersection point p moves on a fixed line x=4.

The elliptic c equation x 2 4 + y 2 3 = 1, the left and right endpoints are a(-2,0), b(2,0).

c = (a 2-b 2) = 4-3) = 1 with the right focus of f(1,0).

Let the straight line through the right focal point be y=k(x-1).

Substituting the elliptic equation gives x 2 4+k 2(x-1) 2 3=1

Sorted out to 3x 2+4k 2(x-1) 2-12=0

3+4k^2)x^2-8k^2x+4k^2-12=0 (1)

Solving this one-dimensional quadratic equation yields two solutions of x, x1 and x2;

At the same time, there are x1+x2=8k 2 (3+4k 2), x1x2=(4k 2-12) (3+4k 2).

Substituting y=k(x-1) can solve two solutions of y, y1=k(x1-1), y2=k(x2-1).

Let the intersection of the line and the ellipse be m(x1,y1),n(x2,y2).

Then the equation for the straight line am is: y=y1 (x1+2)*(x+2).

The equation for the straight line bn is: y=y2 (x2-2)*(x-2).

The intersection p can be solved from two straight lines: y1 (x1+2)*(x+2)=y2 (x2-2)*(x-2).

y1/y2=(x1+2)/(x2-2)*(x-2)/(x+2)=(x1-1)/(x2-1)

x-2)/(x+2)=(x1-1)/(x2-1)*(x2-2)/(x1+2)

1-4/(x+2)=1-(x1+3x2-4)/(x1x2-x1+2x2-2)

x+2)/4=(x1x2-x1+2x2-2)/(x1+3x2-4)

x=4(x1x2-x1+2x2-2)/(x1+3x2-4)-2

x=4[2x1x2-3x1+x2]/(2x1+6x2-8)

2x1x2-3x1+x2)-(2x1+6x2-8)

2x1x2-5(x1+x2)+8

2*4(k^2-3)/(3+4k^2)-5*8k^2/(3+4k^2)+8

8[5k^2-k^2+3]/(3+4k^2)+8

2x1x2-3x1+x2)=(2x1+6x2-8)

2x1x2-3x1+x2)/(2x1+6x2-8)=1

x=4 substituting the solution yields y=6y1 (x1+2)=2y2 (x2-2).

That is, the abscissa of the intersection point is x=4, and the ordinate changes with the change of the point m,n, that is, the straight line x=4

The intersection point p moves on a fixed line x=4.

1. According to the question, e= 3 2=c a a =b +c 2a*2b 2=4 a>b>0

Then a=2 b=1 c= 3 so the equation for the ellipse is x 4+y =1.

2. Set point b coordinates (x0, y0).

Then the coordinates of the midpoint M of AB are (-1+x0 2,y0 2).

a(-2,0), vector ab=(x0+2,y0),q(0,y1),vector qm=(1-x0 2,y1-y0 2)

Vector qa=(-2,-y1), vector qb=(x0,y0-y1)

From the inscription, AB is perpendicular to qm, (x0+2)(1-x0 2)+y0*(y1-y0 2)=0 --1).

qa*qb=4, i.e.: (-2x0)-y1*(y0-y1)=4 --2).

In addition, b is on the ellipse, and there is: x0 4 + y0 = 1 --3).

From (3): x0=4-4y0 -4).

Substituting (4) into (1) gets: y0*(3y0+2y1)=0

When y0=0, x0=2, we get y1=plus/minus 2 2.

When y0 is not equal to 0, y1 = -3y0 2 --5).

Substituting (5) into (2) gets: 15y0 4-4=2x0 --6).

From (3) and (6), we get: x0 = -2 (rounded) or x0 = -2 15, y = plus or minus 2 66 15

So y1 = plus or minus 66 5.

In summary, y1 = plus or minus 2 2 or plus or minus 66 5 (4 solutions in total).

e=ca=root3 2

Rhomboid area 2ab = 4

Find a=2, b=1

So x 2 4 + y 2 1 = 1

I'm sorry, but I don't know!!

Have you ever studied polar coordinates? Take the left focal point f1 (-1,0) as the pole, the positive x direction as the polar axis, and the elliptic equation as p (this is soft, and can not be played out with p instead) = ep (1-ecos), where e=c a, p=a2 c-c

s=1 2*2*sin *p(a)*p(b) (substituting the polar diameter length of ab and calculating it).

Get (tan 2) =1, turn back to Cartesian coordinates, and get the straight line equation x-y+1=0 or x+y+1=0 The above answer is very complete, so I won't say much about it here.

Solution: Let b (x1

y1，c（x2

y2, ellipse.

x220y216

The right focal point of 1 is (2,0).

point a(0,4), and the right focus of the ellipse f2

It's the center of gravity of ABC.

x1+x2+03

2，y1+y2+43

x1x26，y1y2

x 1220

y 1216

1，x 2220

y 2216

Subtract the two formulas to obtain:

x1+x2)(x1−x2)20

y1+y2)(y1−y2)16

Substitute for :

y1−y2x1−x2

That is, the slope of the straight line l is k=

The straight line L passes through the midpoint of BC (3,-2).

The equation for the straight line l is y+2=

x-3) so the answer is 6x-5y-28=0

1. Let m(x,y) and the straight line l:y=k(x 2 3), then the straight line and the ellipse x 12 y 4=1 are used to stand together, and the parameters k are used to represent x and y, and then the parametric equation of the trajectory of the moving point m is obtained, and the ordinary equation is obtained by removing the parameters;

2. Let m(x,y), the right focus is f(2 3,0), and set a(x1,y1), b(x2,y2). Then because a and b are on the ellipse, they are substituted to get:

x1)²/16＋(y1)²/4=1

x2)²/16＋(y1)²/4=1

Subtract the two formulas to obtain:

x1＋x2)(x1－x2)]/16＋[(y1＋y2)(y1－y2)]/4=0

Since x1 x2=2x, y1 y2=2y, and (y1 y2) (x1 x2) is the slope of ab, that is, the slope of mf = y (x 2 3).

After substitution, we get: 4y x(x 2 3)=0 [this is the trajectory equation of point m].