1. locus of the mid-point

question: Avariable line passing through the point (5,0) intersects the lines 3x-4y=0 and 3x+4y=0 at H and K respectively. Find the equaion of the locus of the mid point of HK.

here is my solution:
let P (x,y) be the mid-point
y/(x-5)=m , where m is the slope of variable line
y = mx-5m
put y = mx-5m to 3x-4y=0,
3x-4(mx-5m)=0
x=20m/(4m-3)
put y = mx-5m to 3x+4y=0,
3x+4(mx-5m)=0
x=20m/(4m+3)
since P is the mid-point of HK
x-coordinate of P = 0.5[20m/(4m-3)+ 20m/(4m+3)]
=(80m^2)/(16m^2-9)
put m=y/(x-5)
x=80[y-(x-5)^2)]/[16(y/x-5)^2-9]
90x^3-90x^2+80y^2 -16xy^2+225x=0 is my equation of the locus
but the answer is my textbook says it is 16y^2=9x(x-5)

here is the solution of my textbook:
let H be (4s,3s), K be (4t,-3t).
Slope of HK = (3s+3t)/(4s-4t) = [3(3+t)]/[4(s-t)]
it also equals 3s/(4s-5)
therefore: [3(3+t)]/[4(s-t)]=3s/(4s-5)
8st=5(s+t)
let P(x,y) be the mid -point of HK,
x=(4s+4t)/2=2(s+t) ------------------ (1)
y=3(s-t)/2 -------------------------- (2)
eliminating s and t from (1)and(2):
(3x)^2-(4y)^2 = [6(s+t)]^2 - [6(s-t)]^2
16y^2 = 9x(x-5)

i cannot figure out how can it get (3x)^2-(4y)^2 = [6(s+t)]^2 - [6(s-t)]^2 in order to do the elimination of s and t (seems it could do it just because it knows the final ans.) it also use a different method from mine.
furthermore i want to know my solution is wrong, can anyone point it out?

2. Bump

Bump

3. Originally Posted by afeasfaerw23231233
9x^3-90x^2+80y^2 -16xy^2+225x=0 is my equation of

16y^2 = 9x(x-5)
Note that
$9x^3-90x^2+80y^2 -16xy^2+225x = (x - 5)(9x^2 - 45x - 16y^2)$

Now you tell me why your solution method fails to produce an answer for x = 5.

-Dan

4. the question is to ask the equation of the locus of the mid-point of HK.
so this means my answer 9x^3-90x^2+80y^2 -16xy^2+225x=0 is the same as my book's 16y^2 = 9x(x-5)?
and i also want to know if my method if correct, though it seems tedious and slow while compared to the one in my book's answer.

5. Originally Posted by afeasfaerw23231233
the question is to ask the equation of the locus of the mid-point of HK.
so this means my answer 9x^3-90x^2+80y^2 -16xy^2+225x=0 is the same as my book's 16y^2 = 9x(x-5)?
and i also want to know if my method if correct, though it seems tedious and slow while compared to the one in my book's answer.
The only thing I see wrong with your method (and I would have done it the same way you did) is that it doesn't generalize to the case where the variable line is vertical. This is the origin of the $x \neq 5$ restriction in your solution compared your book's answer.

As to why your equation is so much more complicated I can't really say.

-Dan