Quzhou 20 14 senior high school entrance examination mathematics

Guo Dunqing replied:

20 13 & amp； #8226; Quzhou) In the plane rectangular coordinate system xOy, a rectangle OABC intersects with point AB through the bisector of origin O and points A (0 0,2) and C (6 6,0), and ∠AOC. Point P starts from point O and moves along the direction of ray OD at a speed of ∠ 2 unit lengths per second; At the same time, point Q starts from point O and moves along the positive direction of X axis at a speed of 2 unit lengths per second. Let the moving time be t seconds.

(1) When point P moves to point D, find the value of t at this time;

(2) When t is a value, △PQB is a right triangle;

(3) It is known that the analytical formula of parabola passing through O, P and Q is y =-( 1/t) (x-t) 2+t (t > 0). Ask whether there is a certain moment t, and after rotating △PQB around a certain point by180, the corresponding three vertices just fall on the parabola. If it exists, find the value of t; If it does not exist, please explain why.

Original analysis:

(3) There is such a t value for the following reasons:

Rotate △PQB around a point 180, and the corresponding three vertices all fall on a parabola, then the center of rotation is the midpoint of PQ, and the quadrilateral PBQB'' is a parallelogram.

PO = PQ, from P(t, t) and Q(2t, 0), the coordinate of the rotation center can be expressed as (3 t/2, t/2).

∫ The coordinate of point B is (6,2), and the coordinate of point B is (3t-6, t-2).

Substituting y =-( 1/t) (x-t) 2+t (t > 0) gives 2t2- 13t+ 18=0.

Solution: t 1=9/2, t2=2.

When t 1=9/2, P(t, t)=P(4.5, 4.5), Q(2t, 0) = q (9,0),

When t 1=2, P(t, t) = p (2,2), q (2t,0 0) = q (4 4,0),

The original analysis result is correct, but it is inconsistent with the graphic display result.

What's the problem? Because the number given is wrong, answer the question first, and then give the correct number.

If the center of rotation is the midpoint K in PQ, then the positions of points P and Q are interchanged, point B rotates to b ′, b and b ′ are symmetrical to k, bQB ′ P is a parallelogram, points P and Q are still on the parabola Y =-( 1/t) (x-t) 2+t, and b ′ is also on the parabola; Otherwise, after rotating 180, P and Q will rotate to P ′ and Q ′, which are not on the parabola and do not meet the requirements. This is the center of the question to be answered.

∫ The coordinates of three points are, b (6 6,2), q (2t,0) and p Q(2t, t).

Let the coordinates of k be K(3 t/2, t/2) and the coordinates of point b' be b' (3t-6, t-2).

Substitute the coordinate values of b' (3t-6, t-2) into y =-( 1/t) (x-t) 2+t (t > 0) to get 2t2- 13t+ 18=0.

Solution: t 1=9/2, t2=2.

When t=2, P(t, t) = p1(2,2), q (2t,0) = q1(4,0),

K(3 t/2，t/2)= K 1(3， 1)，B′(3t-6，t-2)= B′(0，0)，

B' coincides with the origin o (0 0,0). P1(2,2) is the vertex of the first parabola,

The first parabola is y =-(1/t) (x-t) 2+t =-(1/t) x2+2x,

y =-( 1/2)x2+2x；

When t&; #8322; =9/2，P(t，t)= P &； #8322; (4.5, 4.5), question and answer; #8322; (2t，0)=Q 1(9，0)，

K(3t/2，t/2)= K & amp； #8322; (6.75，2.25)，B′(3t-6，t-2)= B′(7.5，2.5).

P1(2,2) is the vertex of the second parabola,