Find points of tangency for a given function. I have spent 3 hours or so trying to tackle this problem but the answers I get arent the ones written on the page , I do not want to assume that the asnwers on the page are incorrect so here I go , 
The Problem
The graph :
$y=\frac{5x}{4} + \frac{1}{x}$
(In the first quarter) , has a tangent in point A , and a tangent in point B (both tangent to the function graph) that are perpendicular to each other . (90 deagrees).
It is given that the rate of x of point A is bigger by times 3 than point B (Xa = $t$ while Xb = $\frac{1}{3}t$).


*

*Find the credentials (x , y) of point A , and point B if it is given that the X of point A , is a natural number (bigger than zero , and not a fraction).

*(Didnt finish 1. so didnt get to it)
What I did :
I wrote that the x of point a is t , then I found the derivative of 
$f{(t)}$ which is equal to the angle of tangence (or in other words $m$ of $t$) , derivative : $f'{(t)} = 1.25 + -\frac{1}{t^2}$
after that I found the derivative of point B $(\frac{1}{3}t , f{(\frac{1}{3}t)})$ Which was $f'(\frac{1}{3}t) =  \frac{5}{12} + 3 == 3.416$
I was taught that $m1 * m2 = -1$ . so I took the two m's and followed through :
Ma = $1.25 + -\frac{1}{t^2}$ , Mb = $3.416$ .
Then Ma * Mb == -1 (Right ?)
$(1.25 + -\frac{1}{t^2}) * 3.416 =  -1$
Which led me to $5.27t^2 = 3.416$ led me to this --> $t = 0.7726$
The answers in the book show : A(2,3) and B(0.666 , 2.666) , 
I tried finding my mistake  , but it seems that this also leads me to this. what did I miss here?
Would love some help and insight , The problem was translated , took me a while , hope you understand it (my English isn't perfect to say the least).
 A: The statement $f′(\frac{t}{3})=5/12+3 = 3.416$ is not true. $f′(\frac{t}{3}) = 1.25 - \frac{9}{t^2}$. At $B(2/3,8/3)$, $t=2$, which gives $f′(\frac{2}{3}) = -1$ and $f(\frac{2}{3})=\frac{7}{3}$ (I think there is an error in text book, since it says $\frac{8}{3}$ instead).
At $A(2,3)$: $\;t=6$, and $f′(\frac{6}{3}) = 1$, which is perpendicular as required. $f(\frac{6}{3}) = 3$, a whole number as required.
A: We have that $a=3b$ and $f'(a)\cdot f'(b)=-1$, where $f'(x)=\frac{5}{4}-\frac{1}{x^{2}}$.
Then, $f'(a)\cdot f'(b)=f'(a)\cdot f'(\frac{1}{3}a)=(\frac{5}{4}-\frac{1}{a^{2}})\cdot (\frac{5}{4}-\frac{1}{(\frac{1}{3}a)^{2}})=\frac{25}{16}-\frac{25}{2a^{2}}+\frac{9}{a^{4}}$
Therefore, $\frac{25}{16}-\frac{25}{2a^{2}}+\frac{9}{a^{4}}=-1$. Solving this equation, you'll find that $a=2$ is the only solution that verifies the hypothesis.
Afterwards, $f(a)=f(2)=\frac{5}{4}\cdot 2+\frac{1}{2}=\frac{5}{2}+\frac{1}{2}=3$ and $f(b)=f(\frac{2}{3})=\frac{7}{3}$.
So, you obtain $A(2,3)$ and$B(0.666,2.333)$.
