数学Paper 1讨论专区

Log

回复 2479# whyyie

他应该忙着神鬼传奇。。。

peaceboy

回复 2479# whyyie


    在教小孩子补习，打工

peaceboy

回复 2481# Log


    米有时间玩了

797

回复  797


   
differentiate x^2/2x-1 with respect to x and hence evaluate ∫x(x-1)/(2x-1)^2  ...
peaceboy 发表于 3-1-2011 11:46 PM

whyyie & peaceboy 谢谢你们哦~第2题是我写错了，不好意思~~谢谢你们了~~

peaceboy

版主啊，破百页了你还不要关帖？

yingchin

The straight line L1 passes through A(4,0) and B(2,4) intersects the y-axis at point p. The straight line L2 is perpendicular to L1, and passes through B. It L2 intersects the x-axis and the y-axis at Q and R respectively, show that PR : QR = square root 5 : 3

谢谢

Minute

Find the equation of the locus of midpoint of the chords of the curve X^2+4Y^2=1
with gradient 1.

peaceboy

回复 2487# Minute


Find the equation of the locus of midpoint of the chords of the curve X^2+4Y^2=1
with gradient 1.

    x^2 + 4y^2 =1

2x + 4(2y)(dy/dx) =0
dy/dx = -2x / 8y
         = -x/4y
dy/dx = 1
-x/4y =1
x= -4y


(-4y)^2 + 4(y)^2 =1
16y^2 + 4y^2 =1
y^2 = 1/20
y= +-(1/20)^(1/2)

X^2+4Y^2=1
x^2 + 4/20 =1
x^2 = 16/20
x = +- (4/5)^(1/2)

{-(4/5)^(1/2) , (1/20)^(1/2) }
{(4/5)^(1/2) ,-(1/20)^(1/2) }

m= [(1/20)^(1/2)+ (1/20)^(1/2)]/[-(4/5)^(1/2)-(4/5)^(1/2)]
  = - 1/4

y - (1/20)^(1/2) = -1/4(x+(4/5)^(1/2) )
                        = -x/4  - (1/20)^(1/2)
y=-x/4

Log

回复 2488# peaceboy
find equation of locus of midpoint of chord....with gradient 1
就是 chord of the curve with gradient 1, 不是tangent to the curve with gradient 1..so 我觉得您的做法好像不可以。
(etc 这儿 dy/dx=...)

peaceboy

回复 2489# Log


    最理想的做法不是这样啦，我没时间想
所以想shortcut的

想象chord的midpoint会形成一条直线
所以最简单的可以找到三个点
一个是ellipse 的cneter 0,0
另外两个就是我所找的，gradient =1
三个点作直线就是midpoint

Log

回复 2488# peaceboy
我使用elipse 的 parametric coordinates 和以上的资料来solve...找到 locus y= -x/16

peaceboy

回复 2491# Log


    就写你的step - -

Log

回复 2490# peaceboy

可是dy/dx is gradient fuuction for find gradient of tangent to the curve...但是问题给的gradient 是给chord的

peaceboy

回复 2493# Log


    画图，最小的chord就是 tangent

Minute

The normal to the curve xy=75 at the point (3,25) meets the asymptotes at P and Q.Find the length of PQ.

peaceboy

回复 2495# Minute


The normal to the curve xy=75 at the point (3,25) meets the asymptotes at P and Q.Find the length of PQ.

    xy=75
asymptotes = x-axis and y-axis

x(dy/dx) + y = 0
dy/dx = -y/x
at point (3,25)
dy/dx = -25/3

gradient of normal , m1*(-25/3) = -1
m=3/25

equation of normal
y-25 = 3/25 (x-3)
y= 3x/25 -9/25 + 25
y = (3x-9+625)/25
     =(3x+616)/25

x=0 , y=616/25
y=0 , x= -616/3

P(0,616/25)
Q(-616/3 , 0)
distance =[ (616/25)^2 + (616/3)^2 ]^(1/2)
            = 206.806

很不美的答案 不懂有没有做错

Allmaths

回复  Minute


The normal to the curve xy=75 at the point (3,25) meets the asymptotes at P and Q ...
peaceboy 发表于 20-1-2011 01:58 PM

   
distance =[ (616/25)^2 + (616/3)^2 ]^(1/2)
            =616[634/5625]^(1/2)
            =(616/75)(634)^(1/2)

要美就这样。。。

long_sign

对不起麻烦可以教教一下.........

a)        Prove that if the line lx + my + n =0 touches the ellipse b^2 x^2 + a^2 y^2= a^2 b^2 , then a^2 l^2 +b^2 m^2 = n^2

b)c)        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.

我真的做到哭了......麻烦帮忙一下......

peaceboy

回复 2498# long_sign


   
a)        Prove that if the line lx + my + n =0 touches the ellipse b^2 x^2 + a^2 y^2= a^2 b^2 , then a^2 l^2 +b^2 m^2 = n^2

lx + my + n =0
y=(-n-lx)/m---#

b^2 x^2 + a^2 y^2= a^2 b^2---@
sub # into @
b^2 x^2 + a^2 [(-n-lx)/m]^2= a^2 b^2---@
b^2 x^2 + a^2[(n^2 +2nlx+l^2 x^2)/m^2 ] = a^2 b^2
m^2 b^2 x^2 + a^2n^2 +2a^2 nlx + a^2 l^2 x^2 =m^2 a^2 b^2
(m^2 b^2 +a^2 l^2)x^2 + 2a^2 nlx + a^2n^2 - m^2 a^2 b^2 =0
b^2-4ac = 0
(2a^2 nl)^2 - 4 (m^2 b^2 +a^2 l^2)(a^2n^2 - m^2 a^2 b^2)=0
4a^4 n^2 l^2 - 4 [m^2 b^2 a^2 n^2 - m^4 a^2 b^4 + a^4 n^2 l^2 -a^4 l^2 m^2 b^2] = 0
a^4 n^2 l^2 -m^2 b^2 a^2 n^2 + m^4 a^2 b^4 - a^4 n^2 l^2 + a^4 l^2 m^2 b^2 =0
-m^2 b^2 a^2 n^2 + m^4 a^2 b^4 + a^4 l^2 m^2 b^2=0
- n^2  + m^2b^2 + a^2l^2 =0
a^2 l^2 +b^2 m^2 = n^2 (proven)

下面那题等我做工回来才做
不然等别人解

peaceboy

回复 2498# 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.

问题有米有打错
这样都可以直接找到m是什么了 - -
哪里需要 quadratic equation in m
还是你答案post上来研究看
好累