condition for two curves to have a common tangent Consider the curves $by=x^2+x+\frac{b}{24}$ and $bx=y^2+y+\frac{b}{24}$ if both the curves have a common tangent then the what  values b can take ?
i observed  that  both curves are inverse of each other or they are symmetrical w.r.t to $y=x$ .
My intuition is that this is the key to the problem But i am not able to go further.
So i tried using the general formula for tangent to a parabola of the form $(y-f)^2=4a(x-g)$ which is $y-f=m(x-g)+\frac{a}{m}$ but the expression becomes too complicated.Finally i tried drawing a rough sketch to get some idea but i have to consider too many sub cases.
I came across this problem in a mock test that expects you to solve the problem within 2 minutes.So i guess there is a relatively simple method to approach this problem. Any help will be appreciated .
Edit I agree with all the answers given below.But when i looked at the solution i have no clue of what they are trying to do.i understand how they got $b=\frac{3}{2} and \frac{2}{3}$.I dont understand the rest of the part it is as follows.
let $a=\frac{1}{b}$  if the 2 curves intresect at P1 and P2  but at P1 tangent to first curve is perpendicular to y=x so it is tangent to second curve at P1.
slope of tangent=$2ax+a$ .As (a,x) satisfies this
$2ax+a=-1$ and solving with $x=ax^2+ax+\frac{1}{24}$
$a=\frac{-13+\sqrt{601}}{12}=\frac{1}{b}$
i have written the answer as given .It looks quite absurd .
 A: The parabolas are about the line $y=x$ (they are mirror image of each other about the line $y=x$). So their common tangents will also be symmetric aboy $y=x$. Two possibilities arise:
Case 1: $x=y$ is a common tangent:
If they have to have one common tangent then by putting
$y=x$ in one of them gives $$x^2+(1-b)x+b/24=0$$, this quadratic needs to have only one real root, so the condition $B^2=4AC$ needs to be satisfied:
$$(1-b)^2=\frac{b}{6} \implies 6b^2-12b+6=0 \implies b=\frac{13\pm 5}{12} \implies b=\frac{3}{2},\frac{2}{3}.$$ Only when $b=3/2, 2/3$ $y=x$ is the common tangent.
Case 2: when $x+y=-k$ is the common tangent (more general)
Then we put $y=-k-x$ in the first parabola to get $$x^2-(1+b)x+bk+b/24=0$$
For tangency we demand $B=4AC$, we get
$$(1+b)^2=4bk+b/24 \implies k=\frac{(1+b)^2}{4b}-\frac{1}{24}~~~~(*)$$
Therefor for any real value of $b$ $x+y=-k$, will a common tangent to these parabolas whe $k$ comes from $(*)$.
Case 3: two common tangents
Interestingly, $b=3/2,2/3$ gives $k=1.$ So $x+y=1$ and $y=x$ are two common tangents to the given two parabolas.
See the figures below for $b=4$ (one common tangent, $x+y=73/48.$) and for $b=3/2$ (two common tangents $y=x, x+y=1$).


A: Consider the straight line of equation
$$x+y=c.$$
We intersect it with the parabola
$$by=x^2+x+\frac b{24}$$ and by eliminating $y$ we get a quadratic equation. The discriminant is
$$(b+1)^2-\frac b6+4bc$$
and it cancels (double root) when
$$c=\frac1{24}-\frac{(b+1)^2}{4b}.$$
This correspond to a tangent to the first parabola, and by exchange of $x,y$, it is also tangent to the second one. Hence there is a common tangent for all $b\ne0$.
F.i., with $b=4$,

