The analytical formula of 1 and parabola is:

Y = (- 1/2) x+4x。

2. When t = 4, the line segment eg is the longest, and its length is 2.

3. From tan∠PAE = PE/AP = BC/AB and BC=4, AB=8, we get:

PE/AP = 1/2

∴ PE = AP/2 = t / 2

In Rt△APE, AP = t, PE = AP/2 = t/2,

It is easy to get from Pythagorean theorem: AE = (√ 5/2) t.

∴ EC = AC - AE

= 4√5-(√5/2) tons

There are "three" moments in the movement of points P and Q.

So that △CEQ is an isosceles triangle, which is mainly discussed as follows:

Time ①: When EQ = EC.

Pass point e as EN ⊥ CQ to point n,

* eq = EC，EN ⊥ CQ

Point n is the midpoint of line segment CQ.

At Rt△CEN,

CN = CQ/2 = t/2，EC = 4√5 - (√5/2) t，

In Rt△CAD,

CD = 8，AC = 4√5，

From Rt△CEN ∽ Rt△CAD:

CN :CD = EC :AC

Then (t/2): 8 = [4 √ 5-(√ 5/2) t]: (4 √ 5)

∴ (t/2)×(4√5)= 8 × [ 4√5 - (√5/2) t ]

Multiply both sides by √5 to get:

10t = 160 - 20t

t = 16/3

Time ②: When ②:QE = QC.

Let QM ⊥ EC pass through point Q at point M, then m is the midpoint of line segment EC.

In Rt△CMQ,

CM = CE/2 = [ 4√5 - (√5/2) t ]/2，CQ = t

In Rt△CDA,

CD = 8，CA = 4√5，

From Rt△CMQ ∽ Rt△CDA, we get:

CM :CD = CQ :CA

∴ [ 4√5 - (√5/2) t ]/2 : 8 = t : (4√5)

∴ [ 4√5 - (√5/2) t ]/2 × (4√5)= 8t

∴ 40 - 5t = 8t

∴ 13t = 40

∴ t = 40/ 13

Time ③: When EC = CQ.

∫EC = 4√5-(√5/2)t，CQ = t

∴ 4√5 - (√5/2) t = t

∴ (√5/2) t + t = 4√5

∴ [(2 + √5)/ 2 ] × t = 4√5

∴ t = 4√5 ÷ [(2 + √5)/ 2 ]

= (8√5)/ (2 + √5)

Note: 1. For the solution of time ① and time ②, besides using similar schemes,

Of course, "trigonometric function" can also be used to solve the proportion problem of line segment length;

2. Strongly appeal to all middle school front-line teachers and colleagues to reduce the amount of homework!

Really "elaborate" and extend the variables of the topic in class.

Explore and study with students, and basically "think when you see the topic".