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?
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.
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 $$y = x +2 \text{ with contact points } (-1, 1) \text{ and } (2, 4).$$
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.
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. $
