Find all the parameters and such that the line $y = ax + \frac{1}{2}a - 2$ intersects the hyperbola $xy = 1$ at right angles in at least one point . Problem:
Find all the parameters and such that the line
$y = ax + \frac{1}{2}a - 2$ intersects the hyperbola $xy = 1$ at right angles
in at least one point.
My work:
We try to find tangent to hyperbola such that tangent line intersects given line at rigth angle. We know that two line $y_1=ax+b$ and $y_2=cx+d$ intersects at right angle iff $a=-\frac{1}{c}$.
 So we need to find tangent to hyperbola. Let $t...y=kx+l$ be our tangent line. To find that first we will need $y'$ and we will find it by implicit differentiation.
$$xy=1$$
$$y+xy'=0 \iff y'=\frac{-y}{x}$$
$$k=-\frac{y}{x}$$
But I haven't idea how to finish my work. 
 A: It would be better to use the functional equation
$$y=\frac 1x$$
The derivative is
$$y'=-\frac 1{x^2}$$
So at $x_0=r$ we get $y_0=\dfrac 1r$, $f'(x_0)=-\dfrac 1{r^2}$. For the perpendicular line, we want the slope to be $m=-\dfrac 1{f'(x_0)}=r^2$. Putting that into the point-slope equation of a line, the perpendicular line is
$$\begin{align}
y&=m(x-x_0)+y_0 \\
 &=r^2(x-r)+\frac 1r \\
y&= (r^2)x+\left(\frac 1r-r^3\right)
\end{align}$$
Your question wants that to be $y=ax+\frac 12a-2$ so we see that $a=r^2$ and $\dfrac 12a-2=\dfrac 1r-r^3$. Substituting,
$$\frac 12r^2-2=\frac 1r-r^3$$
That can be simplified to the equation
$$2r^4+r^3-4r-2=0$$
Factoring the left hand side,
$$(2r+1)(r^3-2)=0$$
This has the solution $r=-\dfrac 12$ or $r=\sqrt[3]2$. Since $a=r^2$, this gives the final answer

$$a=\frac 14 \quad\text{or}\quad a=\sqrt[3]4$$

We try that in a grapher, and we see it checks. The two magenta points are the points of intersection where the lines and graph are perpendicular. By "coincidence" one of those points is also a second point of intersection with another line: that can be ignored.

Of course that is not really a coincidence. Your line equation can be rewritten as
$$y=a\left(x--\frac 12\right)+-2$$
which means the question was asking for the lines through the point $\left(-\dfrac 12,-2\right)$, which is on the graphed hyperbola, that are perpendicular to the hyperbola. We get two lines, one for each branch of the hyperbola, and of course both lines go through that point and one is perpendicular to the hyperbola at that point.
A: Let $ (x_1, y_1) $be the point of intersection note that 
$ y_1 = a x_1 - 1/2 a - 2 $ and $ y_1 x_1 = 1 $. the slope at the point $ (x_1, y_1) $ of the line is $ a $ for the hyperbola $ y' = \frac{-1}{(x_1)^2}$ hence $ a = (x_1)^2$ three equations with the variables $ a, x_1,  y_1$
A: Let $P=(x_P,y_P)$ the intersection point between the line and the hyperbola.
The hyperbola has equation $y=\frac{1}{x}$ so the derivative (without implicit differentiation) is $y'=\frac{-1}{x^2}$ and, at $P$ it is $y'(x_P)=\frac{-1}{x_P^2}$.
This is the slope of the tangent to the hyperbola at $P$, and if we want that it's orthogonal to the straight line $y=ax+\frac{1}{2}a-2$ we must have:
$\frac{-1}{x^2}=\frac{-1}{a} \Rightarrow a=x_P^2 $
Now, we want also that $P$ is a point of the straight line, se we must also have:
$\frac{1}{x_P}=ax_P+\frac{1}{2}a-2=x_P^3+\frac{1}{2}x_P^2-2 \Rightarrow x_P^4+\frac{1}{2}x_P^3-4x_P-2=0 \Rightarrow (2x_P-2)(x_P^3-2)=0$ 
Solving this equation you find  the possible values for $x_P$ and  for $a$.
