How to find the common tangent to the curves $y^2=8x$ and $xy=-1$? 
How to find the common tangent to the curves $y^2=8x$ and $xy=-1$ ?

My approach: I used the formulae for tangents of a parabola and hyperbola.For any conic section if $y^2$ is replaced by $yy_1$,$xy$ is replaced by $\frac{
xy_1+yx_1}{2}$,$x$ is replaced by $\frac{x+x_1}{2}$.
I formed two equations and found the condition for their equality but it evaluates to:The common tangent is $y+2x=2$.But that is the wrong answer.
What is the right method?
 A: The problem is that the common tangent may touch the curves in different points.
The tangent to the hyperbola at the point $(t,-1/t)$ has equation
$$
y+\frac{1}{t}=\frac{1}{t^2}(x-t)
$$
that can be rewritten as
$$
x=t^2y+2t
$$
Such a line will be tangent to $y^2=8x$ if the equation
$$
y^2=8(t^2y+2t)
$$
has coincident solutions; the equation is
$$
y^2-8t^2y-16t=0
$$
and its discriminant is $64t^4+64t=64t(t^3+1)$. Since $t\ne0$, the only solution is $t=-1$, so the common tangent is
$$
x=y-2
$$
You can also do it by determining the tangents with the general formula; the tangent to the hyperbola at the point $(t,-1/t)$ is
$$
\frac{-x\dfrac{1}{t}+ty}{2}=-1
$$
or, as determined above, $x-t^2y-2t=0$.
The tangent to the parabola at the point $(s^2/8,s)$ is
$$
8\frac{x+s^2/8}{2}-sy=0
$$
or $8x-2sy+s^2=0$. The two lines must be the same; multiply the first equation by $4$:
$$
\begin{cases}
8x-8t^2y-16t=0\\
8x-2sy+s^2=0
\end{cases}
$$
so $s^2=-16t$ and $-8t^2=-2s$. This implies $16t^4=-16t$, so $t=-1$ as before.
A: by looking at the graphs of $$y^2 = 8x, \, xy = -1$$ i see that it may be possible for a line with positive slope touch $xy = -1$ in the second quadrant and $y^2 = 8x$ in the first quadrant. 
i will pick a point $(a, -1/a)$ on the hyperbola the slope of the tangent at that point is $1/a^2.$  therefore the tangent line has the equation 
$$y + \frac1a=\frac1{a^2}\left(x-a\right)\to y = \frac x{a^2} -\frac2a $$ this line cuts the parabola at $$\left(\frac x{a^2} -\frac2a \right)^2 = 8x  \to (x-2a)^2=8a^2x\to x^2 -4(2a^2+a)+4a^2=0$$ for this quadratic to have a double root, we need the discriminant $$16a^2(2a+1)^2-16a^2 = 0\to 2a+1 = \pm 1\to a = 0, a = -1. $$
we reject $a = 0$ and pick the point $a = -1.$ therefore the common tangent to the curves is $$y = x +2  \text{ with contact points } (-1, 1) \text{ and } (2, 4).$$
A: Find the general equation of the tangent to each curve (call them $\Gamma_1$, $\Gamma_2$) at any point $(x_0,y_0)$.
We have: $\Gamma_1: y^2=8x$
Then, $2yy'=8$, that's $y' = 4/y$ for $y \neq 0$. For $y=0$, the tangent is $y=0$. Then, the slope of the tangent call it $T_1$ at $(x_0,y_0)$ is $4/y_0$. 
$$(T_1): y - y_0 = \frac4{y_0} (x - x_0)$$
As $(x_0,y_0) \in \Gamma_1$, $y_0^2 = 8x_0$.
Replacing this up, and letting $t=y_0$, we have the family of all tangents to $\Gamma_1$.
We do the same thing with $\Gamma_2$
Identify the two family of tangents with each other to get a value of $t$, then replace it in any of the two equations to get the equation of the common tangent.
A: Using differential calculus in the plane:
Differentiate equation of parabola, $ y^{'} = 4/y $
Differentiate equation of hyperbola,$ y^{'} = 1/x^2 $
Eliminate derivative which is same at common tangent and introduce suffix CT for common tangent:
$ y _{CT}= 4 x _{CT}^2 $ which when introduced into parabola equation
$ x _{CT}^4 =8  x _{CT} $ which give $ x_{CT}= (0,2) $
which when substituted into either curve gives $ y_{CT}= (-\infty,4) $
Thus a tangency point on each curve from which a tangent drawn is common for both curves  is 
$ (0, -\infty ) $ and $ ( 2,4)$ 
If equation of tangent is required we can equate the derivative at required point
that gives 
$ y_{CT} = m_{CT} +2 $  and y-axis $ x=0. $
