数学Paper 1讨论专区

long_sign

回复 2500# peaceboy

thank for (a)厡来是要用y=mx+c 来sub 进去
b)的没有打错啊......答按是7m^2 +12m +5 =0, outside

真的不好意事要你们帮忙......

peaceboy

回复 2501# long_sign


   If m is the gradient of the tangent from the point (3,2) to the ellipse 9x^2+16y^2=144, , find a quadratic equation in m .by nothing whether the roots of this equation are real or imaginary ,determine if the point (3,2) lies within the ellipse.

原来是要这答案

9x^2+16y^2=144 ----1
18x + 32 y dy/dx = 0
dy/dx = -9x/16y
m= -9x/16y ----@

m= (y-2)/(x-3)----#

-9x/16y =  (y-2)/(x-3)
-9x(x-3) = 16y (y-2)
-9x^2 + 27x = 16y^2 - 32y
16y^2+9x^2 = 27x + 32y ----2

1-2 , 144= 27x + 32y
x= [(144-32y)/27] ----*

sub * into @
m= -9 [(144-32y)/27] /16y
   = (32y-144)/48y
48ym = 32y-144
y(48m-32) =-144
y=[-144/(48m-32)] -!



sub * into #
m= (y-2)/((144-32y)/27-3)
((144-32y)/27-3)m = y-2
m(144-32y-81)/27 = y-2
m(63-32y)/27 = y-2
m(63-32y)= 27y - 54

sub y=[-144/(48m-32)]
m(63-32(-144/(48m-32))])= 27(-144/(48m-32))] - 54
63m - 32m[-144/(48m-32)] = -3888/(48m-32) - 54
[63m(48m-32) + 4608m] / (48m-32) = [-3888 -54(48m-32)]/ (48m-32)
[63m(48m-32) + 4608m] = [-3888 -54(48m-32)]
3024m^2 - 2016m + 4608m = - 3888 - 2592m + 1728
3024m^2 + 5184m +2160 = 0
7m^2 +12m +5 =0

够力笨长下

9x^2+16y^2=144
(3,2)
LHS = 9x^2+16y^2
      =9(3)^2+16(2)^2
      = 81 + 64
      =145
not equal to RHS
therefore
the point (3,2) not lies within the ellipse.

Lov瑜瑜4ever

S=(2+1)(2^2+1)(2^4+1)(2^8+1)...(2^1024+1)+1

Find S^(1/512)

whyyie

回复 2503# Lov瑜瑜4ever

要想想... 这是STPM水准?

Lov瑜瑜4ever

回复  Lov瑜瑜4ever

要想想... 这是STPM水准?
whyyie 发表于 1-2-2011 08:18 PM

这个是OMK的题目。。。

whyyie

回复 2505# Lov瑜瑜4ever

我就在想嘛...STPM出这题肯定登报纸....

Lov瑜瑜4ever

回复  Lov瑜瑜4ever

我就在想嘛...STPM出这题肯定登报纸....
whyyie 发表于 1-2-2011 08:23 PM

哈哈。。。为什么登报纸。。。？

HanQing

S=(2+1)(2^2+1)(2^4+1)(2^8+1)...(2^1024+1)+1

Find S^(1/512)
Lov瑜瑜4ever 发表于 1-2-2011 07:14 PM

http://www.scribd.com/doc/48001922

Lov瑜瑜4ever

http://www.scribd.com/doc/48001922
HanQing 发表于 2-2-2011 08:19 AM

感激不尽。。。我知道我错哪里勒。。。
我一直把2+1看成2^0+1。。。
所以只算到2^2047+2^2045+2^2043+...+2^1这些单数项罢了==

Lov瑜瑜4ever

Find the positive integers p,q and n such that p and q are prime numbers and
(1/p)+(1/q)+(1/pq)=1/n
我找到2set答案：(p,q,n)=(2,3,1) or (3,2,1)
还有其他组的答案吗？
如果没有的话又要怎样prove除了这2组的答案以外就没有其他的答案了呢〉？

peaceboy

回复 2510# Lov瑜瑜4ever

number theory的要放去数学逻辑那里比较适合
我都没什么读的

Lov瑜瑜4ever

回复  Lov瑜瑜4ever

number theory的要放去数学逻辑那里比较适合
我都没什么读的
peaceboy 发表于 3-2-2011 12:03 AM

噢噢噢。。。可以给我link吗？

peaceboy

回复 2512# Lov瑜瑜4ever


    http://cforum.cari.com.my/forumdisplay.php?fid=140

Lov瑜瑜4ever

回复  Lov瑜瑜4ever
peaceboy 发表于 3-2-2011 12:13 PM

ok。。。tq

jennykuwn

各位大大请给一点提示，这题想了很久还是不会。
ABCD is a square;A is the point(0,-2) and C the point (5,1), AC being a diagonal.Find the coordinates of B and D.

peaceboy

回复 2515# jennykuwn


    我做过类似的题目

http://cforum.cari.com.my/viewthread.php?tid=1513746&page=101&authorid=214702

请看第50楼

Minute

HELP ME ~~
∫(2sin x + 3cos x )/(3sin x + 2 cos X) dx

long_sign

可以给一点提示吗?
1)find the equations of tangent and normal to parabola y^2=4x at the point (t^2,2t).if p is the point with t=√2,find the coordinates of the point q where the normal at the point p meets the parabola again.if o is the origin ,show that op is perpendicular to oq.
2)If pq is a chord which is a normal at point (a,2a) ,which lies on the parabola y^2=4ax,find the length of pq.find also the area of the triangle formed by the chord and the tangents at p and q.
3)The line y-2x+4a=0 intersects the parabola y^2=4ax at the point p(ap^2,2ap) and the point q(aq^2,2aq).find the value of p+q and pq. Hence ,find the coordinates of the mid-point of pq.
Answer:1)ty=x+t^2 ,tx+y=t^3 +2t ;2)8√2a,32a^2 ;3)pq=-2,p+q=1

peaceboy

回复 2517# Minute


    题目加减有没有打错？

peaceboy

回复 2518# long_sign


    1)find the equations of tangent and normal to parabola y^2=4x at the point (t^2,2t).if p is the point with t=√2,find the coordinates of the point q where the normal at the point p meets the parabola again.if o is the origin ,show that op is perpendicular to oq.

y^2 = 4x
2y dy/dx = 4
dy/dx = 2/y
        = 2/ 2t
        = 1/t
y-2t = (1/t)(x-t^2)
ty - 2t^2 = x-t^2
ty=x+t^2

normal = -t
y-2t = (-t)(x-t^2)
y-2t = -tx+t^3
tx+y=t^3 +2t

sub t=√2 into tx+y=t^3 +2t  get 1st equation
y^2=4x << second equation



from equa 1 and 2 , find point q

to show that op is perpendicular to oq.
find the gradient of op and oq , m1m2 = -1 then shown

自己做拉，很累


其他题等别人解
没有我这两天都不得空解