According to the conditions of the topic, it is literally translated into the geometric relationship about the moving point, and then simplified by using the relevant formulas of analytic geometry (the distance formula between two points, the distance formula from point to straight line, the included angle formula, etc.). ). That is, this relationship is transformed into an equation containing x and y, and the trajectory equation of the curve is obtained.
Example: (06 country I) In the plane rectangular coordinate system, there is an ellipse with sum as the focus and eccentricity. Let the part of the ellipse in the first quadrant be a curve C, the moving point P is on C, and the intersection of the tangent of C at point P and the axis is A and B, respectively, and the vector.
Find: the trajectory equation of point m;
Solution: The elliptic equation can be written as: y2a2+x2b2 = 1, where a >；; B>0, and A2-B2 = 33a = 32 leads to A2 = 4 and B2 = 1, so the equation of curve C is: x2+Y24 =1(x >; 0，y & gt0).y = 2 1-x2(0 & lt； x & lt 1) y '=- 2x 1-x2
Let P(x0, y0), because p is on c, use 0.
Y =-4x0y0 (x-x0)+y0。 Let A(x, 0) and B(0, y), x= 1x0, y= 4y0 be given by the tangent equation.
The coordinate from OM→= OA →+ OB→M is (x, y), the equation of C is satisfied by x0 and y0, and the trajectory equation of point M is:
1x 2+4 y2 = 1(x & gt； 1，y & gt2)
Second, the substitution method (correlation point method)
In some problems, it is not convenient to list the conditions that a moving point meets with equations, but a moving point moves with another moving point (called a related point). If the conditions satisfied by the relevant points are obvious or analyzable, then we can use the coordinates of the moving point to represent the coordinates of the relevant points, and according to the equations satisfied by the relevant points, we can get the trajectory equation of the moving point. This method of finding trajectory is called correlation point method. This method is very common and has been tried in the college entrance examination for several years.
Example 2 (Country 03) As shown in the figure, draw the vertical line from the point Q on the hyperbola, with the vertical foot N, and find the trajectory equation of the midpoint P of the line segment QN.
Analysis: from the meaning of the question, the relevant point of the moving point P is Q, and Q is on the hyperbola.
Exercise, so this problem is suitable for using the relevant point method.
Solution: Let the coordinates of the fixed point P be (x, y) and the coordinates of the point Q be (x 1, y 1).
Then the coordinates of point n are (2x-x 1, 2y-y 1).
∫N x+y = 1 on the straight line,
∴2x—x 1+2y—y 1=2 ①
PQ is perpendicular to the straight line x+y=2.
∴ x-y+y 1-x 1 = 0 ②
At the same time ① ② Solve ③.
Q is on the hyperbola, ∴ ④.
③ Substituting into ④, the trajectory equation of the fixed point P is
Third, the definition method.
If the trajectory of the moving point meets the definition of the known curve, the equation can be set first, and then the basic quantity can be determined to find the trajectory equation of the moving point.
Example 3: (Guangzhou No.2 Model in 2005) The moving circle M passes through the fixed point P (-4,0) and is tangent to the circle C: x2+y2-8x = 0, and the trajectory equation of the moving center M is found.
Analysis: according to the meaning of the question || MC |-| MP | = 4, it shows that the absolute value of the distance difference between point M and fixed points C and P is a constant, so the trajectory of point M is a hyperbola. So the definition method is suitable for this problem.
Solution: According to the meaning of the question || MC |-| MP | = 4, it shows that the absolute value of the distance difference between point M and fixed points C and P is a constant, so the trajectory of point M is a hyperbola.
2a=4，∴a=2，c=4，∴b=
Therefore, the trajectory equation of moving center m is
Fourth, the parameter method
Sometimes it is difficult to find the geometric conditions that a moving point should meet, and there is no obvious correlation point, but it is easy to find (or can be found through analysis) that the movement of this moving point is often restricted by another variable (angle, slope, ratio, intercept or time, etc.). ), that is, the coordinates of the moving point (x, y) change with the change of another variable, and we can call this variable as a parameter to establish the parametric equation of the trajectory.
When choosing parameters, we should first fully consider various factors that restrict the moving point, and then choose appropriate parameters. Because the parameters are different, the amount of calculation will be different. Commonly used parameters include angle, straight line slope, horizontal and vertical coordinates of points, line segment length, etc.
Example 4. Parabola y2 = 4px(p & gt；; The vertex of 0) is two perpendicular chords OA and OB. Find the trajectory equation of point P in AB.
Analysis: the moving point P is the midpoint of AB. How to connect P with A and B? While A and B move on parabola, the main condition is that OA and OB are vertical. This problem is suitable for parametric method.
Solution: Let the slope of OA be k > 0.
Then get a () from the solution.
Get b () from the solution.
Let the midpoint of AB be P(x, y), then
The locus equation of the midpoint p obtained by eliminating k is
V. Rail Transit Law
When finding the trajectory of moving point, sometimes there will be a trajectory problem that needs two moving curves to intersect. This kind of problems often get the coordinates (including parameters) of the intersection point by solving the equations, and then eliminate the parameters to find the trajectory equation. This method is usually used in combination with the parameter method.
Example 5 (nationwide in 2003) The constant is known. In a right-angle ABCD, AB=4, BC=4, O is the midpoint OF AB, points E, F and G move on BC, CD and DA respectively, and P is the intersection of GE and of (as shown in the figure). Ask whether there are two fixed points, so that the sum of the distances from P to these two points is a constant value. If it exists, find out the coordinates of these two points and this fixed value; If it does not exist, please explain why.
Analysis: the moving point P is the intersection point with the moving straight line of EG, and the two straight lines are
Movement, this characteristic, can be used by crossing the trajectory.
Solution: as shown in the figure, let P(x, y)
According to the meaning of the question, there are A (-2,0), B (2 2,0), C (2 2,4a) and D (-2,4a).
set up
There are e (2 2,4ak), f (2-4k, 4a) and g (-2,4a-4ak).
The equation of the straight line is 2ax+(2k- 1) y = 0 ①.
The equation of the straight line GE is: -a (2k- 1) x+y-2a = 0 ②.
The parameter k is eliminated from ① ②, and the coordinates of the point P(x, y) satisfy the equation 2a2x2+y2-2ay = 0.
Finishing is available.