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?