A high school math conic curve problem, help, thanks!

Solution: Let the abscissa of the intersection points a and b of the parabola and the straight line be x1, x2 simultaneous x 2 = 2py and y = kx+m, respectively: x 2 -2pkx- 2pm=0 from Veda's theorem:

x1+x2 = 2pk a.

Since the difference in distance from point a, b to y axis is 2k and m>0, when k>0 x1+x2=2k b formula.

Simultaneous a, b gives p=1

When k<0, x1+x2=-2k c.

Simultaneous a,c gives p=-1

Contradict the conditions and give up.

In summary, p=1

It is known that the point m(-2,0) and the point p of n(2,0) satisfy the condition |pm|-|pn|= 2 times the root number 2, and the trajectory of the moving point p is the equation 2 for finding cIf a, b are different points on c, and o is the coordinate origin, find the vector oa multiplied by the minimum value of the ob vector.

Labels: oaob, origin, vector.

Answer (1) Let the p coordinates (x, y).

pm|-|pn|= 2 root number 2

Root number [(x+2) 2+y 2] - root number [(x-2) 2+y 2] = 2 root number 2

To simplify: w is a hyperbola.

According to the definition: c = 2, 2a = 2 root number 2, c 2 = a 2 + b 2

b^2=4-2=2

Then the w equation is: x 2 2-y 2 2 = 1(x<0)

2) When the slope of the line ab does not exist, let the equation of the line ab be x x0, where a(x0,),b(x0,

When the slope of the line ab exists, let the equation of the line ab be y kx b, and substitute it into the hyperbolic equation.

Medium, get: (1 k2)x2 2kbx b2 2 0

According to the meaning of the problem, we can see that there are two unequal positive roots of equation 1°, let a(x1,y1),b(x2,y2), then.

Solution|k|>1, again.

x1x2＋y1y2＝x1x2＋（kx1＋b）（kx2＋b）＝（1＋k2）x1x2＋kb（x1＋x2）＋b2＝

To sum up. The minimum value is 2

Dahl. Rate: Solution: 1Because |pm|-|pn|= 2 times the root number 2, the point m(-2,0),n(2,0), then mn=4>2 times the root number 2, so the trajectory is c is the right branch of the hyperbola with the point m(-2,0), n(2,0) as the left and right focus.

then c = 2, a = root number 2, b = root number 2So the equation for c is x 2 2-y 2 2 = let a, b coordinates are (x1, y1), (x2, y2), then the vector oa multiplied by ob vector = x1*y1 + x2*y2. According to geometric analysis, when the vector OA is perpendicular to the ob vector, the value of the vector OA multiplied by the ob vector is the smallest, which is 0

Anonymity. Rate: 11 stick on the upper curve. c = the root equation is: x -y = 2 (x > 0).

Let the equation for ab be . x=ky+t,a(

The simultaneous equation is solved to obtain (k -1) y + 2kty + (t -2) = 0

y1y2=(t^-2)/(k^-2).

y1+y2=-(2kt) (k -1) vector.

1.Let m(x,y) be the curve c'The symmetry point of its relation to p(a,2a,2a) is n(2a-x,4a-y), n on the ridge curve c:y=-x 2+x+2, 4a-y=-(2a-x) 2+(2a-x)+2, i.e., y=x 2+(1-4a)x+4a 2+2a-2,

for c'equations.

①2，x^2-2ax+2a^2+a-2=0,③

c vs. c'Intersect at a and b points, 4=a 2-(2a 2+a-2)=-a 2+a-2)>0,a 2+a-2<0,-22Let a(x1,y1), b(x2,y2), by ,x1+x2=2a, by ,y1-y2=-x1 2+x1+2-(-x2 2+x2+2).

x1-x2)[-x1+x2)+1],k=(y1-y2) nanopenetration(x1-x2)=1-2a,(The value range of -2k is (-1,5).

It's a little easier to plot You can try to make a (x1,y1), b(x2,y2), and a points of symmetry with respect to p as a'(x1',y1'), the point of b symmetry with respect to p is b'(x2',y2')

Because the state is a, and b is on c, therefore, y1, y2 can be represented by x1, x2, and because of symmetry, x1',y1'It can be represented by x1 and acacia a, x2',y2'It can be represented by x2 and a, and a'Substituting the equation y=-x2+x+2 gives a new equation about x1, because x1 has a solution, according to b 2-4ac>=0, find the range of a, k = (y2-y1) (x2-x1) = 1 - (x1 + x2).

No one answered this question. I'll say a few words.

The two fixed points are symmetrical with respect to the origin, and the product of the distance from the moving point to the two fixed points is equal to the square of the distance from the moving point to the origin. Obvious equiaxed hyperbola. (This question is within a circle).

c = 2, a = b = root number 2. Equation x 2-y 2 = 2The minimum value of the range is the square of the distance from the hyperbolic vertex to the origin = 2.

The moving point is within the circle, and the maximum point is the square of the distance between the intersection point and the origin point above the circle. The square of the distance from the upper intersection to the origin = 16 + root range 16 + root 27.

Have you noticed the range of y?

Write f1p=m, f2p=n, f1f2=2c, and get (2c) 2=m 2+n 2-2mncos 60 degrees from the cosine theorem, i.e.

4c^2=m^2+n^2-mn。

Let a1 be the major semi-axis of the ellipse, and a2 be the real semi-axis of the hyperbola, defined by the ellipse and hyperbola, to obtain m+n=2a1, m-n=2a2, i.e., .

m=a1+a2, n=a1-a2, and substitute them and eccentricity into the previous equation to obtain a1 2-4a1a2+a2 2=0, which can be obtained.

a1=3a2，e1*e2=(c/a1)*(c/a2)=[(c/a2)^2]/3=1，e2=√3，e1=(√3)/3。

The elliptic focal distance sum is 2a, and the hyperbolic difference is 2a,,, in this process the cosine theorem is used, and the hyperbolic separation rate is e,。。 Then the ellipse is the reciprocal, pay attention to one point, they are the same as c I used the ,,, on my mobile phone to write those processes are too troublesome, I will say it in Chinese, I hope it will help you.

Let a(x1,y1)b(x2,y2), bring in 3x 2+4y 2=12 and subtract it.

There are 3(x1+x2)+4(y1+y2)*k'=0 k'=-1 4 Let the midpoint p(x0,y0)6x0-2y0=0 y0=3x0

The midpoint is on the straight line y=4x+m y0=4x0+m x0=. y0=..denoted by m).

Then use p(x0,y0) to find the straight line ab, and use m to represent the simultaneous elliptic equation.

If the discriminant formula "0" can be found to find the range of m value, if the m value cannot be found, it means that there is no symmetry.

If the intersection point is a(x1,y1)b(x2,y2) and the coordinates of the midpoint are (x0,y0), then the linear equation of ab can be set to y=-1 4x+b

x1^2/4+y1^2/3=1①

x2^2/4+y2^2/3=1②

y1=-1/4x1+b③

y2=-1 4x2+b , get. x1-x2)(x1+x2) 4+(y1-y2)(y1+y2) 3=0,get. y1-y2=-1 4(x1-x2) substitute y1-y2 as a whole into the above equation and extract the common factor (x1-x2).

x1-x2)(2x0 4+-1 4*2y0 3)=0Since x1 is not equal to x2, therefore, 1 2 x0-1 6y0=0 and y0=4x0+m

The solution yields x0=-m y0=-3m

x0^2/4 +y0^2/3<1

m^2<4/13

So, -2 13 13