(1) Write the coordinates of point A and the length of AB;
(2) When points P and Q move for several seconds, Q with point Q as the center and PQ as the radius is tangent to line l2 and Y axis, and the value of A at this time is found.
Test center: a function synthesis problem; The nature of tangent; Similar triangles's judgment and nature.
Special topic: geometric moving point problem; Classified discussion.
Analysis: (1) According to the intersection of the linear function image and the coordinate axis, the coordinates can be obtained separately;
(2) According to similar triangles's judgment, we get △APQ∽△AOB. When △ q is tangent to the right Y axis and △ q is tangent to the left Y axis, we get the answers respectively.
Solution: Solution: (1)∵ The image of a linear function is a straight line l 1, and l 1 intersects the X axis and the Y axis at points A and B respectively.
∴y=0，x = ? 4，
∴A(﹣4,0),AO=4，
∫ The coordinates of the intersection of the image and the Y axis are: (0,3), BO=3,
∴ab=5；
(2) From the meaning of the question: AP=4t, AQ=5t, = = t.
∠PAQ=∠OAB，
∴△APQ∽△AOB，
∴∠APQ=∠AOB=90，
Point p is on l 1
∴⊙Q keeps tangent to l 1 during the movement,
(1) When ⊙Q is tangent to the Y-axis on the right side of the Y-axis, let l2 and ⊙Q be tangent to F, and get from △APQ∽△AOB:
∴,
∴pq=6；
If QF is connected, QF=PQ, which consists of △QFC∽△APQ∽△AOB.
Have to,
∴,
∴,
∴QC=，
∴a=OQ+QC=OC=，
② When ⊙Q is tangent to the Y-axis on the left side of the Y-axis, let l2 and ⊙Q be tangent to E, from △APQ∽△AOB: =,
∴PQ=，
If QE is connected, QE=PQ is represented by △QEC∽△APQ∽△AOB: =,
∴=,=,
∴QC=,a=QC﹣OQ=，
∴a value is the sum,
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