1. Prove the problem of "two line segments are equal";
2. The problem of "the maximum length of a moving line segment parallel to the Y axis":
Because the horizontal coordinates of each point on the line segment parallel to the Y axis are equal (always T), with the help of the analytical formula of the function image where the two ends are located, the vertical coordinates of the two ends are represented by algebraic expressions containing the letter T. Then, according to the height of the two endpoints, the length of the moving line segment is expressed as a quadratic decomposition function with the independent variable t and the opening downward by using the formula for calculating the length of the line segment parallel to the Y axis, and the maximum length of the moving line segment can be obtained by using the properties of the quadratic function.
3. Solving the coordinate problem of the known point about the symmetrical point of the known line;
First, use the point oblique method (or K-point method) to find the analytical formula of a straight line passing through a known point and perpendicular to the known straight line, then find the coordinates of the intersection of two straight lines, and finally use the midpoint coordinate formula.
4. Question "Is there a point on the parabola that makes the distance to the fixed straight line maximum":
(Method 1) First, the slope of the fixed straight line is found, from which the analytical expression of the straight line parallel to the fixed straight line and tangent to the parabola can be set (note that the slope of the straight line is equal to that of the fixed straight line, because the slope (k) of the parallel straight line is equal). Then, the equations are formed by the analytical expressions of straight lines and parabolas, and the letter Y is eliminated by method of substitution, and a quadratic equation with one variable about X is obtained, entitled △ =-4ac. There is only one intersection point, so -4ac=0), so you can get the analytical formula of tangent, and then combine the analytical formula of tangent with the analytical formula of parabola into an equation to get the values of X and Y, which are the coordinates of tangent point, and then use the distance formula from point to straight line to calculate the distance from tangent point to straight line, which is the maximum distance.
(Method 2) This problem is equivalent to the maximum area of the corresponding moving triangle, so that when the triangle gets the maximum area, the coordinates of the moving point can be obtained, and then the maximum distance can be obtained by using the distance formula from the point to the straight line.
(Method 3) Firstly, the independent variable is derived from parabolic equation, and the geometric meaning of derivative is used. When the derivative is equal to the slope of a fixed straight line, the coordinates of the obtained point are the points that satisfy the meaning of the question, and the maximum distance can be easily obtained by using the distance formula from point to straight line.
5. Frequently asked questions:
Constant problem in the distance from (1) point to straight line;
Question "Is there a point on the parabola that makes its distance to a fixed straight line equal to a fixed constant?" ;
Firstly, with the help of parabolic analytical formula, the coordinates of moving point are expressed by a letter, and then the equation is established by using the distance formula from point to straight line. By solving this equation, the abscissa of the dispatch point can be found, and then the ordinate of the dispatch point can be found by using the parabolic analytical formula, so as to find the coordinates of the dispatch point on the parabola.
(2) Constant problem in triangle region:
Question "Is there any point on the parabola that makes the area of the moving triangle formed by it and the fixed line segment equal to a constant?" ;
First, find out the length of the fixed line segment, then mark the distance from the dispatching point (whose coordinates need to be expressed in letters) to the fixed line, and then establish an equation with the area formula of the triangle. By solving this equation, the abscissa of the dispatching point can be found, and then the ordinate of the dispatching point can be found with the analytical formula of the parabola, and the coordinates of the moving point on the parabola can be found.
6. Question "Is there a point on a fixed straight line (usually the axis of symmetry of a parabola, or the X-axis or Y-axis or other fixed straight lines) that minimizes the sum of the distances to two fixed points":
Firstly, the coordinates of the symmetrical point of any one of the two fixed points about the alignment are obtained, and then the symmetrical point is connected with the other fixed point to get a line segment. The length of the line segment is the minimum distance that meets the requirements in the question, and the intersection of the line segment and the alignment is the point that meets the minimum distance sum, and its coordinates are easy to obtain (using the method of finding the intersection coordinates).
7. The "maximum (maximum or minimum)" problem of triangle perimeter:
The problem of "whether there is a point on a fixed line to minimize the perimeter of the triangle formed by it and two fixed points" (hereinafter referred to as "the problem of one side moving and two sides moving");
Because there are two fixed points, a triangle has one side (its length can be calculated by the distance formula between two points), as long as the sum of the other two sides is the smallest.
8. The maximum triangle area:
(1) "Is there a point on the parabola to maximize the triangle area formed by it and a fixed line segment" (hereinafter referred to as "the problem that one side is fixed and two sides move");
(Method 1) Firstly, the length of a fixed line segment is calculated by using the distance formula between two points; Then, the maximum distance between the moving point and the fixed line on the parabola is obtained by using the method in the above 3. Finally, using the area formula, the bottom sum height of the triangle is 1/2. The maximum area of a triangle can be obtained, and in the process of solving, the tangent point is the point that meets the requirements of the problem.
(2) The intersection point is parallel to the Y axis to find the intersection point with the fixed line segment (or straight line), thus dividing the moving triangle into two basic model triangles. After displaying the coordinates of the moving point,
Then it can be transformed into a quadratic function problem with downward opening to find the maximum value.
(2) the problem of "the largest area of a moving triangle with uniform motion on three sides" (referred to as "the problem of uniform motion on three sides");
Firstly, the moving triangle is divided into two basic model triangles (a triangle with one side on the X-axis or Y-axis, or a triangle with one side parallel to the X-axis or Y-axis, which is called the basic model triangle), and the coordinates of points with the starting point on the X-axis or Y-axis are set. This kind of problem must contain a set of parallel lines, so that it can be concluded that the divided triangle is similar to another triangle in the figure (usually the largest triangle in the figure). The height of a segmented triangle can be expressed by the property of similar triangles (the ratio of the corresponding side is equal to the ratio of the corresponding height). In this way, the open quadratic function relation of triangle area can be expressed, and the corresponding problems can be easily solved.
9. "Is there a point on the parabola that can make the quadrangle formed by it and the other three fixed points the largest?" ：
Because the quadrilateral has three fixed points, it can be divided into the sum of the areas of a moving triangle and a fixed triangle (two fixed points can be connected to get a fixed triangle), so as long as the area of the moving triangle is the largest, the area of the moving quadrilateral is the largest, and the solutions of the maximum area of the moving triangle and the coordinates of the moving point on the parabola are 7.
10, the problem of "determining the area of a quadrilateral";
There are two common solutions:
Scheme (1): connect a diagonal line and divide it into the sum of the areas of two triangles;
Scheme (2): Make a vertex of a quadrilateral that is not on the X axis or Y axis perpendicular to the X axis (or Y axis), or connect this vertex with the origin, and divide it into the sum (or difference) of the areas of a trapezoid (usually a right-angled trapezoid) and some triangles, or the sum (or difference) of the triangle areas of several basic models.
1 1. "Two triangles are similar" problem:
12. Question "Is there a point on the function image that makes it form an isosceles triangle with the other two fixed points";
First, find out whether the point in the problem is the vertex of an isosceles triangle. (If there is a bottom, there is only one situation; If one side is waist, there are two situations; If only three points form an isosceles triangle, there are three situations). Firstly, with the help of the analytical formula of the image where the moving point is located, the coordinates (mother indication) of the moving point are expressed. According to the classification, the equations are established by using the distance between two points and the isosceles formula under the corresponding category. By solving this equation, the abscissa of the moving point can be found, and then the ordinate of the moving point can be found with the help of the functional relationship of the image where the moving point is located. Pay attention to remove the points that are not in the meaning of the question (that is, you can't form a triangle).
13, "Is there a point on the image that makes it form a parallelogram with the other three points":
For this kind of problem, there are at least two fixed points among the four points in the problem. Use the "one parent indicator" of the moving point coordinates to set the coordinates of all other moving points (if there are two moving points, it is obvious that each moving point should choose a parameter letter to the "one parent indicator" of the moving point coordinates), choose a known point as the starting point of the diagonal, and list all possible diagonals (obviously at most three), at this time, the corresponding other diagonal. Find the midpoint coordinates of the two diagonal lines in each case. According to the judgment theorem of parallelogram, the two midpoints coincide and the coordinates are equal and corresponding. List two equations and solve them.
In addition, there are:
(1) What if such a moving point forms a rectangle? Let the moving point form a parallelogram first, and then verify that the two diagonals are equal? If they are equal, the moving point can form a rectangle, otherwise such a moving point does not exist.
(2) What if there is such a moving point to form a prism? Let the fixed point form a parallelogram first, and then verify whether any group of adjacent sides are equal? If they are equal, the moving point can form a prism, otherwise such a moving point does not exist.
(3) What if such a moving point forms a square? Let the fixed point form a parallelogram first, and then verify whether any group of adjacent sides are equal? And the two diagonal lines are equal? If they are all equal, then the moving point can form a square, otherwise such a moving point does not exist.
14. The question "Is there a point on the parabola that causes the multiplication relation of sum and difference between the areas of two graphs": (This is a "single-action problem" (that is, the problem of combining the definite analytic formula with the moving graph), and the following 19 is actually a special case of this type. )
Firstly, the direct moving point coordinates are set by the method of "one-mother indication" of moving point coordinates, and then the expression (if the figure is a moving figure, it can only represent its area) or calculation formula (if the figure is a fixed figure, it can calculate its specific area) is used, and then the equation of the area relationship between two figures is established and solved from the meaning of the question. (Pay attention to remove irrelevant points). If the coordinates of the indirect moving point are found in the problem, then continue to solve after finding the coordinates of the direct moving point.
15. Question "Is there a point on the graph (straight line or parabola) that makes it form a right triangle with the other two fixed points":
If both sides of the right angle are not parallel to the Y axis, first set the coordinates of the starting point (a parent index), calculate the slopes of both sides of the right angle with the slope formula according to the classification of topics, and then use the vertical slope conclusion of two straight lines (no straight line parallel to the Y axis) (the slope of two straight lines multiplied by-1) to get an equation and solve it.
If one of the two sides of a right angle is parallel to the Y axis, the slope formula cannot be used at this time. The remedy is that if the remaining points (points not on the straight line parallel to the Y axis) are perpendicular to the straight line parallel to the Y axis or the perpendicular of the straight line parallel to the Y axis intersects with another related image, the coordinates of the related point can be easily obtained.
16. "Is there a point on the image that makes it form an isosceles right triangle with the other two fixed points?" .
(1) If the fixed point is a right-angled vertex, first use the K-point method to find the analytical formula of the straight line where the other right-angled edge is located (if the slope does not exist, you can directly write the equation of the straight line where the other right-angled edge is located according to the fixed right-angled point), then use this analytical formula and the analytical formula of the image where the point is located to form an equation group, find out the coordinates of the intersection point, and then use the distance formula between two points to calculate the two right-angled edges. If it is equal, the intersection is consistent with the topic, otherwise it is irrelevant and discarded.
(2) If the moving point is a right-angled vertex: first, use the K-point method to find the analytical expression of the vertical line in the middle of the fixed line segment, then use the analytical expression of the image where the point is located to form an equation group, find the coordinates of the intersection point, and then calculate the slopes of the point and the two straight lines where the two fixed points are located, and multiply the two slopes to see if the result is-1. If it is-1, it means that intersection is needed; On the contrary, give up.
17. "The problem contains two equal angles. Find the coordinates of related points or the length of line segments";
The problem contains two equal angles, that is to say, it should be solved by triangle similarity. At this time, finding the basic model "A" or "X" in triangle similarity is the key and breakthrough.