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
Find all values of a such that y=a/(x−8) and y=x^2−16x+64 intersect at right angles?
please how the steps :)
5 answers
I don't get how you solved 2^(-3/2)and got 1/sqrt2
thanks steve :)
2(8+cbrt(a) - 16 = cbrt(a)^2 / a
where did cbrt(a)^2 / a come from??
where did cbrt(a)^2 / a come from??
and how did you go from cbrt(a)^2 / a into 1/cbrt(a) ????