数学Paper 1讨论专区

Allison

1.The line y=mx+c intersects the parabola y^2=4ax at two points P(x1,y1) and Q (x2,y2). If M is the midpoint of PQ, find the locus of M as m varies.

2.Given the A(4,0) and T is a variable point that lies on a circle x^2+y^2 +4y = 12. If P is the midpoint of AT, show that P lies on a circle.

高手帮我一下。。。谢谢！

@影子@

帮我解答下！
Find the perpendicular distance between the parallel lines 6x-8y=15 and 6x-8y=20

Allison

L1: 6x-8y=15
L2: 6x-8y=20
Let (1,-9/8) lies on L2
SD= I 6(1)-8(-9/8)-20 I / sqrt 6^2 + (-8)^2
     = 5/10 = 1/2

不懂对吗？

@影子@

你答对了，谢谢！

Let (1,-9/8) lies on L2

我想问你这个coordinate怎样找出来？

darksider

原帖由 Elsiewang 于 2009/8/28 03:16 PM 发表
+ sign times -sign equals 0？

Anything that is squared will be >0 except (0)^2

(-2)^2 = 4
(2)^2 =4

Therefore (any value except 0)^2 >0

darksider

原帖由 @影子@ 于 2009/9/5 09:49 PM 发表
你答对了，谢谢！

我想问你这个coordinate怎样找出来？

L1 and L2 are parallel. Their gradient, m1 & m2 =3/4
Let L3 be the line perpendicular to both of them.
Since L3 is normal to the parallel lines, m3 x m1 = -1
Therefore, m3 = -4/3

Now that we know L3 has an equation of y = (-4/3)x + C
Since we don't know C yet, we have to find it out.

Now let's pick any point on L1 and L2.
L2 -  6x-8y = 20
When x=0, -8y=20, y = - 5/2
Therefore, (0 , -5/2) is a point on L2

Assume that L3 crosses point (0, -5/2) (can be any point but since we create this L3, we can make it cross any point we like) , therefore, L3 - y = (-4/3)x + C ;  (-5/2) = (-4/3)(0) + C ; C = -5/2

Hence, equation of L3 is y = (-4/3)x -5/2
Now we want to find the intersection between L3 and L1
L1 = 6x-8y=15
-8y = 15 -6x
y = (3/4)x - 15/8

Equating L1 and L3 ; (3/4)x - 15/8 = (-4/3)x -5/2
                                     x = -3/10 ; hence, y = (3/4)(-3/10) -15/8
                                                                      =-2.1
Therefore, intersection point between L1 and L3 is (-0.3 , -2.1)
Intersection point between L2 and L3 is (0, -2.5)

Now, applying distance formula, d = sqrt [ (y2-y1)^2 + (x2-x1)^2]
After working out, d = 1/2 = 0.5

*I'm using normal distance formula to find out the answer

Another way is to use shortest distance formula which is
shortest  perpendicular distance = | ax +by +c| / sqrt ( a^2 + b^2)
L2 - 6x-8y =20 , 6x-8y -20 = 0 ( arranging it to general form - ax + by + c )
Now we apply the (-0.3, -2.1), the point of intersection btw L1 and L3 into that formula
Therefore , using the formula , shortest perpendicular distance = | 6(0) -8(-5/2) -20 | / sqrt (6^2 + (-8)^2)
                                                                                                           = |-5| / 10 = 5/10 = 1/2

Got it? Try to draw the diagram and you will see the intuition behind it.
If you don't know how to use the formula, just use the normal distance formula to find it out.

darksider

原帖由 Elsiewang 于 2009/8/28 03:16 PM 发表
+ sign times -sign equals 0？

(x+2)^2 >0

^2 代表 square... 不是乘。

你们需要数学帮忙的话，可以add [email protected] , 我有空时可以帮你。

Allison

Given the A(4,0) and T is a variable point that lies on a circle x^2+y^2 +4y = 12. If P is the midpoint of AT, show that P lies on a circle.

请问这题怎么做？？

darksider

原帖由 Allison 于 2009/9/6 10:04 AM 发表
Given the A(4,0) and T is a variable point that lies on a circle x^2+y^2 +4y = 12. If P is the midpoint of AT, show that P lies on a circle.

请问这题怎么做？？

Let T = (x , y)
Midpoint of AT = P = ( 2- x/2 , -y/2)

Since P is the midpoint between T and A
Hence, PT = AT
(PT)^2 = (AT)^2
using distance formula, you will get the final equation which is 2x^2 + 2y^2 -8x = 0
This equation is a circle because
i)Coefficients of x and y are same.
ii) no xy term.

Allison

Midpoint of AT = P  不是(x+4)/2,(y+0)/2吗？？
你怎么有减的？
如果照你方法(PT)^2 = (AT)^2做的话，最后2边都会算到一样的东西

darksider

原帖由 Allison 于 2009/9/7 07:09 PM 发表
Midpoint of AT = P  不是(x+4)/2,(y+0)/2吗？？
你怎么有减的？
如果照你方法(PT)^2 = (AT)^2做的话，最后2边都会算到一样的东西

Midpoint P = ( 4-x /2 , 0-y/2)

(AP)^2 = (PT)^2
(sqrt [ (x- (4-x)/2)^2 + ( y-(-y/2))^2])^2 =(sqrt [((4-x)/2 - 4)^2 +(-y/2 -0)])^2

Allison

我记得midpoint是+的，为什么你用减呢？

darksider

原帖由 Allison 于 2009/9/7 09:01 PM 发表
我记得midpoint是+的，为什么你用减呢？

對，我錯了，midpoint是（x1+x2 /2 , y1+y2 /2)

找對的midpoint了，能找到answer？

harry_lim

我想问一下。。。要find the sets of values
如果有一些题目出现modulus的equation
用square both sides的。。。这个答案可以接受吗？

idontwant2b

squaring both side 只可以用在两边都有知道的sign后的，
比如，
|1/x+1|<2,
两边都不是positive的话，就最好不要用，
可能会和答案一样，
可是working会扣分。

lonely_world

Given y=ax&#178;+bx+c,a.b &c are postive constants. Show that the least value of y is -&#8710;/4a where &#8710;=b&#178;-4ac, & find the corresponding value of x. Sketch the graph for (1)&#8710;>0,(2)&#8710;<0

这题怎样做啊?

Log

first,u must know least value of y mean that MIN.VALUE of y
so, 用completing square method to complete this equation,then ,u will get the answer in this form,y=a(x+p)+q)
where p is min value of y and its corresponding x is -p.(form 4的）
For the graph,
D>0,has two real different roots,U-shaped curve that intersect x-axis 2 diff points。
D<0，no real roots ，U shaped curve but does not intersect x-axis
note that both this graphs must sketched in quadrant II , y-intercepts =c.

lonely_world

原帖由 Log 于 9-9-2009 02:43 PM 发表
first,u must know least value of y mean that MIN.VALUE of y
so, 用completing square method to complete this equation,then ,u will get the answer in this form,y=a(x+p)+q)
where p is min value of y a ...

我懂这些..
可是..那个min point好像怪怪的..
最后那两个图是怎样画的?
画出来是怎样的?

Log

mid point ?got problem ma?
i did already ok
shape?i told u already?!

Log

sorry min point not mid point