As shown in the figure, it is known that there is a straight line and a curve in the rectangular coordinate system. The straight line and the positive half axis of x-axis and y-axis intersect at point a and point B respectively, and OA = ob = 1. This curve is a branch of the image of function y = 12x in the first quadrant. Point P is any point on this curve, and its coordinates are (a, b). The perpendicular lines PM and PN made from point P to x-axis and y-axis are m, n, Line AB intersects PM and PN at points E and f respectively (1) Calculate the coordinates of points E and f respectively (use the algebraic expression of a to represent the coordinates of point E, and use the algebraic expression of B to represent the coordinates of point F, only need to write the results, not the calculation process); (2) Calculate the area of △ OEF (the result is expressed by the algebraic formula containing a and b); (3) The values of AF and be are calculated respectively (the results are expressed by algebraic expressions containing a and b); (4) Please prove whether △ AOF and △ BOE are necessarily similar; if they are not necessarily similar or not, briefly explain the reasons. Solutions: (1) point E (a, 1-A), point F (1-B, b); (2 points) (2) S △ EOF = s rectangular monp-s △ emo-s △ fno-s △ EPF, =ab-12a(1-a)-12b(1-b)-12(a+b-1)2, =12 (a + B-1); (4 points) （3）BE=a2+(1-1+a)2=2a, AF = (1-1 + b) 2 + B2 = 2B; (6 points) (4) (7 points) It is proved that: OA = ob = 1, ∴∠FAO=∠EBO； ∵ point P (a, b) is a point on the curve y = 12x, 2 ab = 1, that is AF &; be = 1; OA and ob = 1, ∴AFOB=OABE； Why is 2Ab = 1, that is AF ﹥ 8226; be = 1; Why is 2Ab = 1?

As shown in the figure, it is known that there is a straight line and a curve in the rectangular coordinate system. The straight line and the positive half axis of x-axis and y-axis intersect at point a and point B respectively, and OA = ob = 1. This curve is a branch of the image of function y = 12x in the first quadrant. Point P is any point on this curve, and its coordinates are (a, b). The perpendicular lines PM and PN made from point P to x-axis and y-axis are m, n, Line AB intersects PM and PN at points E and f respectively (1) Calculate the coordinates of points E and f respectively (use the algebraic expression of a to represent the coordinates of point E, and use the algebraic expression of B to represent the coordinates of point F, only need to write the results, not the calculation process); (2) Calculate the area of △ OEF (the result is expressed by the algebraic formula containing a and b); (3) The values of AF and be are calculated respectively (the results are expressed by algebraic expressions containing a and b); (4) Please prove whether △ AOF and △ BOE are necessarily similar; if they are not necessarily similar or not, briefly explain the reasons. Solutions: (1) point E (a, 1-A), point F (1-B, b); (2 points) (2) S △ EOF = s rectangular monp-s △ emo-s △ fno-s △ EPF, =ab-12a(1-a)-12b(1-b)-12(a+b-1)2, =12 (a + B-1); (4 points) （3）BE=a2+(1-1+a)2=2a, AF = (1-1 + b) 2 + B2 = 2B; (6 points) (4) (7 points) It is proved that: OA = ob = 1, ∴∠FAO=∠EBO； ∵ point P (a, b) is a point on the curve y = 12x, 2 ab = 1, that is AF &; be = 1; OA and ob = 1, ∴AFOB=OABE； Why is 2Ab = 1, that is AF ﹥ 8226; be = 1; Why is 2Ab = 1?

It should be y = 1 / 2x
Multiply both sides by 2x at the same time to get 2XY = 1
If (a, b) is taken in, 2Ab = 1
Using Pythagorean theorem to calculate AF = (radical 2) B
Similarly be = (radical 2) a
AF · be = (radical 2) a · (radical 2) B = 2Ab = 1