Current location - Training Enrollment Network - Mathematics courses - Geometric problems of mathematical functions in grade three. ........
Geometric problems of mathematical functions in grade three. ........
Idea: (1) Substitute the coordinates of B and C into the analytical formula of parabola to get the value of undetermined coefficient;

(2) Because the diagonals of the rhombus bisect each other vertically, if the quadrilateral POP'C is a rhombus, then the point P must be on the perpendicular bisector of OC, from which the coordinates of the point P can be obtained, and the coordinates of the point P can be obtained by substituting it into the analytical formula of parabola;

(3) Because the area of △ABC is unchanged, when the area of quadrilateral ABPC is the largest, the area of △BPC is the largest; Taking P as the parallel line of Y axis, intersecting with straight line BC in Q and intersecting with X axis in F, it is easy to get the analytical expression of straight line BC, and the abscissa of point P can be set. Then the length of PQ can be obtained according to the analytical expressions of parabola and straight line BC, and the area of △ABPC can be obtained with PQ as the base and the absolute value of the abscissa of point B as high, thus the functional relationship between the area of quadrilateral ACPB and the abscissa of point P can be obtained.

Process: solution: (1) is obtained by substituting the coordinates of b and C.

Solution:? b=-2

c =-3;

So the expression of quadratic function is:

y=x^2-2x-3

(2) there is a point p, which makes the quadrilateral POPC rhombic; Let the coordinates of point P be (x, x 2-2x-3), and the intersection of PP' and CO in E.

If the quadrilateral POP'C is a diamond, then PC = po.

When PP' is connected, PE⊥CO is connected to E,

∴OE=EC=? 32

∴y=? -32;

∴x^2-2x-3=? -32

X 1=? 2+ root number 10/2, x2=? 2- radical number 10/2 (irrelevant, omitted)

∴ The coordinates of point P are (? 2+ 102,? -32)

(3) A parallel line with Y axis passing through point P intersects BC at point Q and OB at point F, and let P(x, x 2-2x-3),

Easy to get, the analytical formula of straight line BC is y=x-3.

Then the coordinate of point Q is (x, x-3);

S quadrilateral ABPC=S△ABC+S△BPQ+S△CPQ.

=? 12AB? OC+? 12QP? +? 12QP? novio

=? 12×4×3+ 12(-x2+3x)×3

=? -32(x-32)2+758

What time? When x=32, the area of quadrilateral ABPC is the largest.

At this point, what are the coordinates of point P? (32,-154), what is the maximum area of quadrilateral ABPC? 758.

On the left is Tu Tu who asked the first question, and on the right is Tu Tu who asked the second question ~