Find all values of a such that y=a/(x−9) and y=x^2−18x+81 intersect at right angles?

Question

please how the steps :)

Answer 1

y = a/(x-9) = a(x-9)^-1
dy/dx -a(x-9)^-2 = -a/(x-9)^2

y = x^2 - 18x + 81 = (x-9)^2
dy/dx = 2(x - 9)

recall that for perpendicular lines their slopes are negative reciprocals of each other

(x-9)^2/a = 2(x-9) or a = (x-9)^2 / (2(x-9))
a = (x-9)/2

Where do they intersect ?
a/(x-9) = (x-9)^2
a =(x-9)^3

then
(x-9)^3 = (x-9)/2
(x-9)^2 = 1/2
x-9 = ± 1/√2 or ± √2/2

sub into a = (x-9)/2 = ± √2/4

Answer 2

I don't think the above answer is valid because you will continue getting the same answer of ± √2/4 no matter the question.

for example y=a/(x−8) and y=x^2−16x+64, doing it with the same method will receive the same value. I've tried 3 times and continue getting ± √2/4. Unless I've done one of your steps wrong, I don't think this method is valid.