數學測驗唔識...

天才A

AB 係diameter
OM bisect EB
EM 係tangent to 個circle

prove 角DOM=角MOB

呢個係 part a 第一題
已經唔識做.....



呢條差b part
前面a part 搵 axis of symmetry
之後搵返條 方程式

b part
P(-4,0) R(2,0)  (一開始係比 (x+1)^2-9)
If Q is a moving point on the graph and below x axis
S is a point above x axis
PQRS is a parallelogram
A(x) is the area of PQRS
1.find the domain of PQRS
2.find the maximum area of PQRS

-終場ソ使者-

1)
You prove AE//OM
<ADO=<DAO
<DOM=<ADO
<MOB=<DAO
<DOM=<MOB

2.1)
Domain (for x) is -4<x<2
2.2)
To find max of A(x),
just consider the height of tri. PRQ (PR is its base)
since PR is a constant, A'(x) is directly prop. to its height   (A'(x) represents the area of tri. PRQ)
ie find the max height
therefore, A(x)=2A'(x)

天才A

我prove唔到AE//OM...
果時剩5分鐘 就係掙扎係度...
不過今日聽人講先發現可以用 similar triangle AEB OMB....

1)
You prove AE//OM
-終場ソ使者- 發表於 2-12-2011 19:31

第2題解得好好
唔該晒....

我竟然傻到連domain 都寫唔到
救命

-終場ソ使者-

in fact mid-pt thm may be a better for proving AE//OM !

天才A

in fact mid-pt thm may be a better for proving AE//OM !
-終場ソ使者- 發表於 13-12-2011 20:15

!__!
你又岩wow...

不過果時我又真係好蠢
我冇consider 到ABE 呢個三角形....
所以我白白浪費左OB同OA呢兩條r