High and high conic curves are in demand for 20, and high and high conic curves are in demand

1) a 2 + b 2 = 7, a 2-b 2 = 1, a 2 = 4, b 2 = 3, the elliptic equation is x 2 4 + y 2 3 = 1

2) Let d(-4,d),de:y=k(x+4)+d, and substitute the above equation to get 3x 2+4(kx+4k+d) 2=12,3+4k 2) x 2+8k(4k+d)x+4(4k+d) 2-12=0,

64k^2(4k+d)^2-4(3+4k^2)[4(4k+d)^2-12]

4[36+48k^2-12(4k+d)^2]=0,3+4k^2=16k^2+8dk+d^2,12k^2+8dk+d^2-3=0,③

The coordinates of the tangent point e: by ,xe=-4k(4k+d) (3+4k 2), substituting ,ye=k(12-4dk) (3+4k 2)+d=(3d+12k) (3+4k 2), left focus f1(-1,0), vector f1d*f1e=(-3,d)*(3-12k 2-4dk) (3+k 2), (3d+12k) (3+4k 2))).

9+36k^2+12dk+3d^2+12dk)/(3+4k^2)

9+36K 2+24DK+3D 2) (3+4K 2)=0 (by ),F1D F1E, with de as the diameter of the circle C over F1

In the first question, the hyperbola is connected to the straight line to obtain an equation, and the range of values of k can be obtained by solving 0.

In the second question, let a(x1,y1)b(x2,y2) because f connects af and bf in a circle, so af is perpendicular to bf, so kaf kbf 1 gets x1x2 x1 x2 y1y2 0 and then according to the equation obtained in the first question, x1x2, x1+x2After that, y1y2 can be obtained from the linear equation, and an equation about k can be obtained by bringing the obtained result into x1x2 x1 x2 y1y2 0.

You can figure it out yourself!! I'm tired to death, I must adopt it!!

You're too deductible, there's no benefit, and you're going to have someone to help you do your homework (.).

Let me tell you what Zheng Jiaoyu thinks.

1.Let the linear equation y=,kx1-k)q(x2,kx2-k)2The straight line is connected to the ellipse, and the Vedic theorem is used to obtain x1+x2, x1 x2

3.The vector product is expressed as coordinates, and the relation about k, x1, x2, m is obtained.

4.Substituting the rolling formula of the obtained relation in 2 into the relation formula in 3, x1 and x2 are represented by k to obtain m.

You can try it the way I say sails are good. The detailed steps are too troublesome, if you can't figure it out, you can ask me again.

As shown in the figure, it is known that the major axis of the ellipse x 2 a 2+y 2 b 2=1 (a b 0) is ab, and the line l of the crossing point b is perpendicular to the x axis The fixed point through which the straight line (2-k) x-(1+2k)y+(1+2k)=0(k r) passes is exactly one of the vertices of the ellipse, and the eccentricity of the ellipse e= 3 2

1) Find the standard equation for an ellipse;

2) Let p be any point on the ellipse that is different from a and b, PH x-axis, h is the perpendicular foot, extend hp to the point q so that hp=pq, connect aq to extend the intersection line l at the point m, n is the midpoint of mb Try to judge the position relationship between the line Qn and the circle O with ab as the diameter

1) Analysis: The fixed point through which the line (2-k) x-(1+2k) y+(1+2k)=0(k r) passes is exactly one of the vertices of the ellipse.

2-k）x-（1+2k）y+（1+2k）=0==>2x-y+1=k(x+2y-2) *

Let x+2y-2=0==>x=0,y=1;y=0,x=2

Substituting the * equation shows that the point (0,1) is the fixed point through which the line (2-k) x-(1+2k)y+(1+2k)=0(k r) passes.

b = 1 The eccentricity of the ellipse e = 3 2 = (a 2-b 2) a = = > a 2 = 4

The standard equation for an ellipse is: x 2 4 + y 2 = 1

2) Analysis: The major axis of the ellipse is AB, and the straight line L through point B is perpendicular to the X axis.

The straight line l: x=2, and the equation for a circle with ab as the diameter is: x 2 + y 2 = 4

Let p be any point on the ellipse that is different from a and b, then p(2 (1-y0 2),y0).

Through p as the pH x-axis, h is the vertical foot, and the HP is extended to the point Q so that HP=PQ

q(2√(1-y0^2),2y0)

Connect the AQ extension intersection line L at the point m, where n is the midpoint of the MB.

aq：k=y0/[√(1-y0^2)+1]==>y=y0/[√(1-y0^2)+1]*(x+2)

m(2,4y0/[√(1-y0^2)+1])，n(2,2y0/[√(1-y0^2)+1])

qn equation: k=y0[1-1 ( (1-y0 2)+1)] [ (1-y0 2)-1]= y0[ (1-y0 2) ( (1-y0 2)+1)] [ (1-y0 2)-1].

y0√(1-y0^2)/(-y0^2)=-√(1-y0^2)/y0

>y-2y0=-√(1-y0^2)/y0*[x-2√(1-y0^2)]==>y=-√(1-y0^2)/y0*x+2/y0

>x√(1-y0^2)+y0y-2=0

The distance from the origin to the line qn d= |x√(1-y0^2)+y0y-2|/√[(1-y0^2)+y0^2]=2

The straight line qn is tangent to the circle o with ab as the diameter;

(1) The fixed point (0,1) over the straight line (2-k)x-(1+2k)y+(1+2k)=0 is the vertex of the ellipse, b = 1, eccentricity = 3 2, a = 2, the standard equation for the ellipse is x 4 + y = 1

2)hp=pq?hq=ph?

If the landlord has learned the parametric equations, he can do so.

1) From the known:

The equation for a straight line is y= (2-k) (1+2k) +1, and the intersection point of the line with the ellipse is (0,1).

b²=a²-c²=1 a²=4

The standard equation for an ellipse is x 4 + y = 1(2) and the m coordinate is (2,y).

By (1) know the point p(2sin, cos). From the meaning of the title;

q(2sin,2cos),n(2, y2) and aqm collinear,y-2cos2-2cos = 2cos 2sin +2

y=4cos sin +1 , then:

The straight line qn is cos y + sin x - 2 =0 (with a two-point formula, this step is more cumbersome).

The radius of a circle with ab as the diameter is 2, then the distance from the center of the circle to the straight line is d = 2 cos + sin = 2, so the position relationship between the line qn and the circle o with ab as the diameter is tangent.

1. There is a question to know c a=e= 2 2,a= 2c,m(0,b),a(-a,0),b(a,0).

Vector mf=(c,-b),fb=(a-c,0),,(a-c)c= 2-1 with the problem of knowing the vector relation, combined with a= 2c, the solution is c=1, a =2, b =1

The elliptic equation is x 2 y 1

2. Let the found straight line exist, let y=x+m and bring in the elliptic equation to get 3x +4mx+2m -2=0,x1+x2=-4m 3,x1x2=(2m -2) 3,y1y2=(x1+m)(x2+m)=x1x2+m(x1+x2)+m

The product of two vectors = 0, i.e. two straight lines perpendicular. That is, the product of the two slopes = -1, i.e., y1 (x1-1) y2 (x2-1) = -1

That is, y1y2+(x1-1)(x2-1)=0, that is, y1y2+x1x2-(x1+x2)+1=0, and bring the above conclusion into the solution to obtain m= 7 3-2 3

The equation for the straight line is y=x 7 3-2 3

1.Solution: p(-1, 2 2), b(c,0), m(0,y).

The intersection point M of the line segment Pb and the Y axis is the midpoint of the line segment Pb, so 0=(-1+C) 2, C=1

So b 2 = a 2-1, substituting the elliptic equation.

x^2/a^2+y^2/(a^2-1)=1

The point p(-1, 2 2) is on the ellipse, 1 a 2 + (1 2) (a 2-1) = 1

The solution is a 2 = 2, so b 2 = 2 - 1 = 1

So the standard equation for an ellipse is x 2 2 + y 2 = 1

2.Solution: Let the desired straight line exist, let y=x+m and bring in the elliptic equation to obtain 3x +4mx+2m -2=0,x1+x2=-4m 3,x1x2=(2m -2) 3,y1y2=(x1+m)(x2+m)=x1x2+m(x1+x2)+m

The product of two vectors = 0, i.e. two straight lines perpendicular. That is, the product of the two slopes = -1, i.e., y1 (x1-1) y2 (x2-1) = -1

That is, y1y2+(x1-1)(x2-1)=0, that is, y1y2+x1x2-(x1+x2)+1=0, and bring the above conclusion into the solution to obtain m= 7 3-2 3

The equation for the straight line is y=x 7 3-2 3

Ellipse c: (x 2) + y = 1The straight line l y=x+t. can be setSynapsing with the elliptic equation (x 2) + y = 1 gives 3x +4tx + 2(t -1) = 0

8(3-t²)＞0. ∴t²＜3.Set the points p(p,p+t),q(q,q+t), and p+q=-(4t) 3,pq=2(t-1) 3

and point f(1,0),m(0,1), vector fp mq=(p-1,p+t) q,q+t-1)=0 ∴p-1)q+(p+t)(q+t-1)=0.Sorting can obtain 2pq+(t-1)(p+q)+t-t=0.

Substituting the results of Vedic theorem, we can get 3t +t-4=0The solution yields t1=1, t2=-4 3All meet t 3

The straight line l y=x+1, or y=x-(4 3)

You don't seem to have made it clear!

1.Let a(x1,y1), b(x2,y2).

They are on a parabola, so there are: y1 = 2px1, y2 = 2px2

According to the analytic formula of the parabola y = 2px, there must be: x1, x2, x0>0

The parabolic alignment is: x=-p 2

Let the distances from a, m, and b to the alignment be d1, d0, d2

According to the second definition of parabola: the distance from the point on the parabola to the focal point must be equal to the distance to the quasi-line

af|=d1,|mf|=d0,|bf|=d2

af|,|mf|,|bf|into a series of equal differences.

af|+|bf|=2|mf|

|d1|+|d2|=2|d0|

According to the definition of coordinates, we can get: d1=x1+p 2, d0=x0+p 2, d2=x2+p 2 (x1, x0, x2, p are all positive, so you can take off the absolute value symbol).

(x1+p/2)+(x2+p/2)=2(x0+p/2)

x1+x2=2x0 ②

From q(x0+p,0), a(x1,y1),b(x2,y2), we get:

aq|=√bq|=√

aq|^-bq|^=x0+p)-x1]^+y1^-[x0+p)-x2]^-y2^

Substitution:

aq|^-bq|^=x0+p)^-2x1*(x0+p)+x1^+y1^ -x0+p)^+2x2*(x0+p)-x2^-y2^

2x1x0-2px1+x1^+2px1 +2x2x0+2px2-x2^-2px2

x1^-x2^ -2x1x0+2x2x0

x1+x2)(x1-x2)-2x0(x1-x2)

x1+x2-2x0)(x1-x2)

Substitution, get:

aq|^-bq|^=0

|aq|=|bq|

2.|mf|=d0=|x0+p/2|=x0+p/2

x0+p/2=4 ③

by o(0,0),q(x0+p,0).

Friends hail head = >|oq|=|x0+p|=x0+p

x0+p=6 to get: p=4

The parabolic equation is: y = 8x