High 2 curves and equations, high and high curves and equations

The upstairs idea is correct, but in the end there is a problem when describing the center of the circle, because the equation is (x+2a) +y = (2m), that is, the center of the circle is on the negative half axis, and point b is on the positive half axis, how can b be the center of the circle, the center of the circle should be at a point with a distance of 2a to the left of a, the others are correct, in fact, the equation is not so complicated, it can be obtained according to the observation of geometric properties, and c is cf ad, because d is the midpoint of bc, then ad=2m, a is (-2a, 0), No matter how C changes, the above conclusion can be obtained by passing C as CF AD, then the point c is on a circle with the center of the circle (-2a,0) and the radius of 2m, so the trajectory equation is (x+2a) +y =(2m).

With a as the origin, the straight line where ab is located is the positive semi-axis of the x-axis to establish a Cartesian coordinate system, then a(0,0),b(2a,0), let c(x0,y0), then d((x0+2a) 2,(y0+0) 2),|ad|=((x0+2a) 2-0) +y0+0) 2)-0), known |ad|=m, so (x0+2a) 4+y0 4=m, i.e., (x0+2a) +y0 =(2m), due to the arbitrariness of c(x0,y0), replace c(x0,y0) with c(x,y), then the trajectory equation of c is.

x+2a)²+y²=(2m)²。It is a circle with b as the center and 2m as the radius.

High School Curves and Equation Problems.

1.Knowing that the center of the hyperbola is at the origin, a focal point is f(7,0), the line y=x-1 intersects with it at two points m and n, and the abscissa of the midpoint of mn is -2 3, then the hyperbolic equation is ---

2.It is known that the circle x2+y2=1, the point a(1,0), abc is connected to the circle, bac=60°, and when bc moves on the circle, the trajectory equation of the midpoint of bc is ---

3.It is known that a(4,0),b(2,2) are the points in the ellipse x2 25+y2 9=1, and m is the moving point on the ellipse, then |ma|+|mb|maximum value ---

4.If the coordinates of the bottom of the two flagpoles are set as a(-5,0) and b(5,0) respectively, then the trajectory of the point on the ground with the same elevation angle of the pole top is ---

Answer: Lianli.

Straight line y=x-1

x 2 repentance file a 2-y 2 b 2=1

There is (1 a 2-1 b 2) x 2+2x b 2-1-1 b 2=0

There is also a horizontal ordinate of the midpoint of mn with -2 3

So x1+x2 2=a 2 (a 2-b 2)=-2 3 --1)

A focal point is f(7,0) then there is c 2 = a 2 + b 2 = 7 ---2).

Simultaneous (1) (2) yields a 2 = 2 and b 2 = 5

So the hyperbolic pre-circle equation to be found is x 2 2-y 2 5 = 1

First connect the center of the circle ob oc to do ok perpendicular bc to k point.

Since a=60°, boc=120° isosceles triangle obc has a vertex angle of 120°, so ok=

Obviously, the k point is the midpoint of BC, and there is ok=

So the trajectory of the k point is the circle with the center of the circle as the origin and the radius as follows: x 2 + y 2 = 1 4

Draw an ellipse figure by yourself, make an ellipse with left focus f(-4,0) Obviously, point A is the right focus.

The first definition of the ellipse is: MA+MF=2A=10

then MA+MB=2A-MF+MB

In the triangular MFB there is: |mb-mf|≤bf=2√10

So ma+mb=2a-mf+mb [10-2 10,10+2 10].

ma+mb)max=(2a-mf+mb)max=10+2√10

If the point is on a straight line ab, then there are 2 -10 = 3 5-x x = 20

If the point is within [-5,5], you may wish to set the elevation angle to b.

5cotb+3cotb=10

x=5cotb-5

Obviously, x=Actually, if you look at the x-axis, only 2 points meet the conditions.

If you look down at the flagpole from the air, the trajectory of the ground point that meets the conditions is actually two straight lines.

Let the coordinates of the m point be (x,y).

Then the coordinates of point A are (x,0), the coordinates of point B are (0,y), and points A, B, and P are in a straight line.

Then (x-3) (0-4)=(3-0) (4-y) gives (x-3)*(y-4)=12 and x≠3,y≠4.

If |pm|-|pn|=6, then point p is on the extension line of mn, and the trajectory is y=2(x 5).

2. Let any point on the curve C1 be (x,y), then its symmetry point about (-2,1) is (-4-x,2-y), and since this point is on the first trace c of the curve, the equation for curve C1 is 2-y=(-4-x) 2-2(-4-x)+2, that is, y=-x 2-10x-24

Let any point on the curve c2 be (x,y), then its symmetry point with respect to x-y-3=0 is (3+y,x-3), and the point at which the lu qin trembles is on the curve c, so the equation for the curve c2 is.

x-3=(3+y)^2-2(3+y)+2