a maths locus

其實題目是a maths locus, 只需要用中五學生水平所做到的便可, 只是solve 2個unknown根本沒有需要用Cramer’s Rule. 毫無疑問, Andrew兄在數學上的程度較很多人高, 我亦同意真正好的答案，是教方法，不是答問題, 但是我個人認為因應發問者的程度來作答才是最合適, 正如沒有人會對一個請教簡單算術題的小學生長篇大論地講Number Theory.

以下是中五程度的方法, 其實分別不大, 只是應用一些較為平易近人的語言罷了.
Using point slope form, assuming equation of the variable line as:
(y – 0) = m(x – 5), y = mx – 5m … (i)
(可能你會覺得這樣不夠precise, 因為這條equation涵概不到vertical line x = 5)
3x – 4y = 0 … (ii)
3x + 4y = 0 … (iii)

Solving (i) and (ii) for H,
Sub (i) into (ii),
3x – 4(mx – 5m) = 0
(3 – 4m)x + 20m = 0
x = 20m/(4m – 3)
y = m[20m/(4m – 3)] – 5m = 5m[4m – 4m + 3]/(4m - 3) = 15m/(4m – 3)
Therefore H is [20m/(4m – 3), 15m/(4m – 3)]

Similarly, Solving (i) and (iii) for K,
Sub (i) into (iii),
3x + 4(mx – 5m) = 0
(3 + 4m)x – 20m = 0
x = 20m/(4m + 3)
y = m[20m/(4m + 3)] – 5m = 5m[4m – 4m – 3]/(4m + 3) = -15m/(4m + 3)
Therefore K is [20m/(4m + 3), -15m/(4m + 3)]

By Mid-point Theorem, the locus of the mid-point of HK is:
x = [20m/(4m – 3) + 20m/(4m + 3)] / 2 = 20m [4m + 3 + 4m – 3] / 2(4m + 3) (4m – 3) x = 80m²/(4m + 3) (4m – 3) … (iv)
y = [15m/(4m – 3) - 15m/(4m + 3)] / 2 = 15m [4m + 3 – 4m + 3] / 2(4m + 3) (4m – 3) y = 45m/(4m + 3) (4m – 3) … (v)

但通常中五a maths這類題目都不接受parametric form做答案, 所以須去掉parameter m:

(iv) / (v), x / y = 80m / 45 = 16 m/ 9,
m = 9x / 16y … (vi)

Sub (vi) into (v),
y = 45(9x / 16y)/[4(9x / 16y) + 3] [4(9x / 16y) – 3]
y[4(9x / 16y) + 3] [4(9x / 16y) – 3] = 45(9x / 16y)
(9)16y² [3x / 4y + 1] [3x / 4y – 1] = 45(9x)
[3x + 4y] [3x – 4y] = 45x
9x² - 16y² - 45x = 0

The locus required is: 9x² - 16y² - 45x = 0

Assuming the variable line be

ax + by + c = 0 <- this can represent ANY line including vertical and horizontal ones.

since this line passes through (5, 0)

5a + 0b + c = 0
c = -5a

In other words, the line is

ax + by = 5a --- (1)

3x - 4y = 0 --- (2)

3x + 4y = 0 --- (3)

Solving this equation sets (1) with (2):

We skip the mechanics and get (Cramer's rule helps, if you know)

Hx = -20a / (3b - 4a)
Hy = -15a / (3b - 4a)

That will be the coordinate for H.

Again, solving this equation sets (1) with (3):

We skip the mechanics and get (Cramer's rule helps, if you know)

Kx = 20a / (3b + 4a)
Ky = -15a / (3b + 4a)

That will be the coordinate for K.

Finally, we compute the mid-point as:

Mx = (Hx + Kx)/2 = [-20a / (3b - 4a) + 20a / (3b + 4a)]/2
My = (Hy + Ky)/2 = [-15a / (3b - 4a) - 15a / (3b + 4a)]/2

Let's simplify it first:

Mx
= 10a[1/(3b + 4a) - 1/(3b - 4a)]
= 10a[(3b - 4a) - (3b + 4a)]/(9b^2 - 16a^2)
= -80a^2/(9b^2 - 16a^2)

My
= -7.5a[1/(3b + 4a) + 1/(3b - 4a)]
= -7.5a[(3b + 4a) + (3b - 4a)]/(9b^2 - 16a^2)
= -45ab/(9b^2 - 16a^2)

At this stage, if it is acceptable for two parameters parametric form, that probably is the simplest.

If we strike for a single parameter one, we have to assume b =/= 0. Notice, the scaling factor of b has nothing to do with the locus or the straight line, without loss of generality let b = 4/3 to make life easy.

Then
Mx = -80a^2/(16 - 16a^2) = 5a^2/(a^2 - 1)
My = -60a/(16 - 16a^2) = 15a/4(a^2 - 1)

Then we obtains a single parameter parametric form, remember to add the point (5,0) to the set as well since we excluded the vertical line when we assume b =/= 0

There maybe imprecision with the solution, but the general strategy follows.

2007-04-21 16:12:12 補充：
謝謝，你的答案的確比較簡潔。
我也沒這個心情去eliminate parameter。