Finding the equation of conic curve (including finding trajectory) is an important basic knowledge of analytic geometry, and it is also the key content of college entrance examination every year. Its main content is the solution of elliptic, hyperbolic and parabolic equations. The solution of this kind of problems often involves the comprehensive application of functions, inequalities, equations, triangles, straight lines and other related knowledge, the idea of combining numbers with shapes, the idea of functions and equations, and the idea of transformation. Therefore, in the college entrance examination, conic curve is often used as the carrier to comprehensively examine students' comprehensive ability.

2. Common methods for solving conic equation

Definition method, undetermined coefficient method, direct method, substitution method, parameter method, geometric method, etc. The key is the combination of form and number to establish an equal relationship.

3. Learning and examination requirements of this unit

According to the given conditions, we can choose the appropriate coordinate system, find the curve equation and draw the curve represented by the equation.

4. The general steps and key points of solving the curve equation are as follows

Establish system, form, simplify and prove.

There are two situations in the first step of "building a system (establishing a coordinate system)" in practical problems: (1) There is already a coordinate system in the studied problem, and then the equation can be found in the given coordinate system; (2) There is no coordinate system in the condition. At this point, you must first choose the appropriate coordinate system. Usually, the point in a special position is always selected as the origin, and the straight lines perpendicular to each other are the coordinate axes.

The second step is the most important step. We should carefully analyze the characteristics of the curve, pay attention to revealing the implicit conditions, master the equivalence relationship related to any point on the curve, meet the geometric conditions, and list the equations. In the process of transforming geometric conditions into algebraic equations, we should pay attention to the definition of quadratic curve in junior high school and the application of plane geometry knowledge, and also often use some basic formulas, such as the distance formula between two points, the distance formula from point to line, the slope formula of line and so on.

Thirdly, in the process of simplification, we should pay attention to the rationality and accuracy of operation and deformation to avoid "losing solution" and "adding solution".

For the fourth step, there is no requirement (theoretically necessary) in middle school, and there will be no problem in most cases, but in case of special circumstances, it is necessary to explain it appropriately. For example, according to the meaning of the question, although the coordinates of some points satisfy the equation, they are not on the curve, so they can be deleted by limiting the value range of x and y.

5. Example analysis

Example 1 Find the trajectory equation of the center p of a moving circle tangent to the fixed point A (2 2,0) and the fixed line x=-2.

The solution is easy to know, as shown in the figure. The distance from the moving point to the fixed point is equal to the distance from the fixed line. According to the definition of conic curve, the trajectory of moving point is parabola y2=2px, where p = 4. Therefore, the trajectory equation of point P is y2=8x.

Example 2 (1992 national college entrance examination) focuses on f1(-2,0) and F2 (6 6,0), and the equation of hyperbola with eccentricity of 2 is _ _ _ _ _ _ _ _ _ _ _.

The center of hyperbola is (2,0), c=4, c/a=2, a=2, b2= 12.

∴ The curve equation is.

Example 3 (the national subject of1993) If the moving circle and the fixed circle x2+y2= 1 and x2+y2-8x+ 12=0 are all tangent, then the trajectory equation of the center of the moving circle is ().

A. parabola B. circle C. a branch of hyperbola D. ellipse

The solution conditions are O: x2+y2 = 1, r1=1; M: (x-2) 2+y2 = 4, r2=2, m (2 2,0), let the center of the motion circle be P(x, y) and the radius be r, then,

∴,

According to the definition of hyperbola, the central locus of a moving circle is a branch of hyperbola. So choose C.

Example 4 There are three different points A (X 1, Y 1), C (X2, Y2) and B (0 0,6) on the upper branch of the hyperbola, and the distances from them to the focus F (0 0,5) are arithmetic progression, so find the value of y 1+y2.

The directrix-hyperbola of the solution is m:y=5/ 12,

If AA 1⊥m is in AA 1, ∴,

Similarly,

∵,

∴ 2,

∴y 1+y2= 12。

Explanation 1] The above four examples are all solved according to the definition of conic, which is one of the most important solutions to solve the conic equation. Among them, Embodiment 3 and Embodiment 4 use the first and second definitions respectively. In fact, whenever "focus radius (the connecting line between the focus and the point on the curve)" appears in the topic, the definition of conic curve should be considered. If "directrix" appears, the second definition will definitely be used.

2) When the moving circle is tangent to the fixed circle, it is necessary to connect two center points (commonly used auxiliary lines in plane geometry) to find the distance relationship between the center points. Its trajectory is usually one of parabola, ellipse or hyperbola. In this respect, Example 3 is more representative.

Example 5 has the same asymptote as hyperbola, and the equation of hyperbola passing through point A(2, -3) is _ _ _ _ _ _ _ _ _.

The hyperbolic equation is,

Point a is on the hyperbola, Ⅷ

The hyperbolic equation is:

Explain that this topic examines the application of undetermined coefficient method and hyperbolic equation of * * * asymptote system.

Example 6 (1997 National College Entrance Examination) Ellipse C and ellipse are symmetrical about the straight line x+y=0, and the equation of ellipse C is ().

A.B.

C.D.

By analyzing any point M(x, y) of ellipse C, it is easy to know that the symmetrical point of m about the straight line x+y=0 is on the known ellipse, and the equation of ellipse C can be obtained.

Let's solve any point M(x, y) on ellipse C, and use the symmetrical point of m about the straight line x+y=0 as M'(-x, -y). According to the question, m' is a known point on the ellipse.

The equation is,

So choose a.

Example 7 (Guangdong problem of1990) When the moving point moves on the circle x2+y2= 1, the locus equation of the midpoint of the connecting line with the fixed point (3,0) is ().

A.(x+3)2+y2 = 4 b .(x-3)2+y2 = 1 c .(2x-3)2+4 y2 = 1d .(x+3/2)2+y2 = 1/2

The solution is shown in the figure, let m be any point on the circle,

The fixed point is A (3 3,0), even AM, let the midpoint of AM be n, n, and the midpoint of OA be c (3/2,0).

Then CN= 1/2, then the distance from n to c is fixed length 1/2.

Its trajectory equation is (x-3/2)2+y2= 1/4, that is, (2x-3)2+4y2= 1.

So choose C.

It shows that the solution of 8 cases and 9 cases is geometric method, that is, when there are plane figures such as circles and parallelograms. In the topic, we should make full use of their geometric properties and find the geometric conditions that the moving points meet to establish the equivalence relationship. This method is simpler than other methods in this problem.

Example 8 Given the fixed point A (3 3,0), P is the moving point x2+y2= on the unit circle, and the bisector of ∠AOP intersects with PA at m, the trajectory equation of m point is found.

The solution is shown in the figure. Let the coordinates of m and p be (x, y) and (x) respectively. ，y .)

Properties of bisector from triangle angle.

, namely

∴

x= xo=，

y= yo=

∫xo2+yo2 = 1, and the trajectory equation of ∴ m point is () 2+()2= 1

M: (x-+y2 =。

The solution to this problem is method of substitution, that is, the coordinates xo and yo of the moving point on the known curve are expressed by the coordinates X and Y of the moving point on the trajectory, and then the equation of the trajectory can be obtained by substituting it into the equation of the known curve, which is also a highly efficient method to solve the conic curve equation.

Example 9 the discriminant of equation ax2+bx+c=0(a.b.c∈R, a≠0) is equal to 1, and the product of two roots is a constant k(k≠0). Find the curve equation represented by point (b, c).

According to the meaning of the question, there are

b2-4ac= 1，

If A is eliminated, b2-4 is b2-.

The curve equation of point (b, c) is x2-.

It shows that the solution to this problem is parameter method.

Example 10( 1993 college entrance examination questions) is in ⊿PMN with an area of 1, TG ∠ PMN = 1/2, TG ∠ MNP =-2. Establish an appropriate coordinate system and solve the elliptic equation with m and n as the focus and passing through point p.

The solution is shown in the figure, and the coordinate system is established with the straight line of MN as the X axis and the perpendicular line of MN as the Y axis.

Let an elliptic equation with m, n as the focus and p as the intersection point, with m (-c, 0) and N(c, 0) as the focus.

From TG ∠ PMN = 1/2 and TG = (∠ PMN) = 2, the equations of linear PM and PN are y=(x+c) and y=2(x-c) respectively.

X=, y= and y = are obtained by solving two equations simultaneously, that is, the coordinate of point P is (,).

So ⊿ PMN =

C= is obtained from the condition S δ PMN = 1, that is, the coordinate of point P is ().

Substituting into the elliptic equation, it is simplified as 3b4-8b2-3=0.

The solution is b=, a2=b2+c2=3+=.

Therefore, the equation is.

Example 1 1 (1998 national college entrance examination questions) As shown in the figure, the straight line l 1 intersects with l2 at point M, and the electricity Nl 1, and the distance from any point on the curve segment C with A and B as endpoints is equal to the distance to point N. If ?.

The solution is shown in the figure. Establish a coordinate system with l 1 as the X axis and the middle vertical line of MN as the Y axis. According to the meaning of the question, curve segment C is a parabola with N as the focus and l2 as the directrix.

Let the equation of curve c be y2 = 2px (p >; 0), (xAXxB, y>0), where xa xA and xB are the abscissa of a and b, respectively, and p=.

∴M(-p/2,0),N(p/2,0)。

By =，= 3

(xA+p/2)2+2p xA= 17┄①，

(xA-p/2)2+2p xA =9 ┄②

Xa = p/4 is obtained by simultaneous solution of ① and ②, which is substituted into ① formula and expressed by P >; 0，p=4，xA = 1； Or p=2 and xA=2.

∵⊿AMN is an acute triangle, ∴ P/2 > Therefore, p=2 and xA=2 are omitted.

From the point p on the curve segment c, XB =-p/2 = 4 is obtained.

To sum up, the equation of curve segment C is y2=8x( 1≤x≤4, y>0).

Explain that the above two examples mainly examine the basic idea of choosing a suitable coordinate system according to the given conditions and finding the analytic geometry of curve equations (using undetermined coefficient method), and examine the concepts and properties of ellipses and parabolas, the relationship between curves and equations, and the ability to comprehensively apply knowledge.

6. Summary

Solving the equation of conic curve (including trajectory) is the basic content of analytic geometry, and it is necessary to master the application of various methods. Proper selection of solutions, proper selection of coordinate systems, and reasonable and sufficient use of numerical conditions to establish equality relations are the basic skills to solve this kind of problems. The main law of solving problems can be summarized as follows: "The definition of curve must be clearly remembered, the relationship between numbers and shapes must be proved, the coordinate system must be selected, the method should be reasonable, and the process should be smooth. Select parameters, introduce parameters, make good use of parameters, substitute them for elimination, and skillfully transform. The coefficient to be determined is the constant method, and the column equation is the key, clarifying the relationship and ideas, and decrypting the overall situation. "