Find all values of a such that y=a/(x−8) and y=x^2−16x+64 intersect at right angles?

Question

please how the steps :)

Answer 1

The curves intersect where
a/(x-8) = (x-8)^2
a = (x-8)^3

x = 8 + cbrt(a)

Now, we need to find a such that the curves are perpendicular.

slope of x^2 - 16x + 64 = 2x - 16
slope of a/(x-8) = -a/(x-8)^2
perpendicular slope = (x-8)^2/a

So, when x = 8 + cbrt(a), and the slopes are equal, we have

2(8+cbrt(a) - 16 = cbrt(a)^2 / a
16 - 2cbrt(a) - 16 = 1/cbrt(a)
2cbrt(a) = 1/cbrt(a)
a = 2^(-3/2)

cbrt(a) = 1/√2
Checking the slopes at x = 8 + 1/√2

2x - 16 = 2(8+1/√2) - 16 = 16 - √2 - 16 = √2

(x-8)^2/a = 1/2 / 2^(-3/2) = 2^(3/2)/2 = √2

Answer 2

I don't get how you solved 2^(-3/2)and got 1/sqrt2

Answer 3

thanks steve :)

Answer 4

2(8+cbrt(a) - 16 = cbrt(a)^2 / a

where did cbrt(a)^2 / a come from??

Answer 5

and how did you go from cbrt(a)^2 / a into 1/cbrt(a) ????