find the locus of the point $Q$ with ellipse $\frac{x^2}{3}+\frac{y^2}{2}=1$ Let $P(x,y)$ on the ellipse $\Gamma$:$\dfrac{x^2}{3}+\dfrac{y^2}{2}=1$,and the circle $\tau$:$x^2+y^2=1$,if $PA,PB$ tangent the circle $\tau,A,B\in\Gamma$
and $AQ,BQ$ tangent the circle $\tau$.where $Q=AQ\bigcap BQ$
find the locus of the ponit $Q$?
 A: Long story short, it is the ellipse going through $Q$ that shares the four common (complex) tangents of the circle and the outer ellipse.
An equivalent condition to the sharing tangents condition is that after a suitable rescaling, all ellipses share the same foci.

Let $E = \{(x,y) ; (X,Y) \mid x^2+y^2=1 ; 2X^2+3Y^2=6 ; xX+yY = 1 \}$.
It is the set of pairs of points $(P,Q)$ where $P$ is on the circle, $Q$ is on the outer ellipse, and $(PQ)$ is tangent to the circle.
Given one point $P$ on the circle, there are usually two (possibly complex) possible points $Q$, so this is a branched cover of the circle.
Since there is a duality betwen points and lines in the projective plane (or vector lines and vetor planes in space), the set of tangent lines of each curve can be again described as a curve in the projective plane, and since they both have degree $2$, they should have $4$ intersection points, which means that there should be $4$ common tangents to the circle and the ellipse.
If you actually do the computation you find that those tangents are the lines $1 \pm \sqrt 2y \pm ix = 0$ where each sign can be chosen independantly.
Then, $E$ is a two-fold cover of the circle with four branch points. Since the circle has genus $0$, $E$ is a curve of genus $g$ with $2g-2 = 2*(-2)+4=0$ so $g=1$ and $E$ is an elliptic curve after choosing a point on it. 

For any other conic that shares those same tangents with the circle, you can make $E'$ with the same construction, and you will obtain a new curve that is also a two-fold cover of the circle and who has the branch points at the very same places, which means that there is a pair of isomorphisms $E \leftrightarrow E'$. (You can't really single one out in particular)
In fact, since the tangents touch the circle at the points $(x = \pm i, y = \pm \sqrt 2)$, we can use the curve $E_0 = \{(x,y,z) \in \Bbb C^3 \mid x^2+y^2=1; z^2 = 1+x^2 \}$ (because it has the same ramification data as all the others). All of those are isomorphic and between any two of them there are exactly $2$ isomorphisms fixing the circle.

Now we have a bunch of natural involutions that are defined on $E$.
If $(P,Q) \in E$, we can keep $P$ on the same place of the circle and switch $Q$ with the other intersection of the tangent line with the ellipse to get $(P,Q')$. This involution $f$ has $4$ fixpoints corresponding to the $4$ common tangents. Seen in $E_0$ this is the map $(x,y,z) \mapsto (x,y,-z)$.
When $\alpha$ varies along $E$, the divisors $\alpha + f(\alpha)$ are all equivalent to each other.
Next, you can look at the dual situation, you keep $Q$ on the outer ellipse and switch $(PQ)$ with the other tangent line through $Q$. It again has $4$ fixpoints (since it is the dual problem, they occur at the $4$ intersection points of $E$ with the circle, where there is only one tangent and not two).
Now this involution $g$ depends on which conic you are using, but then again, in terms of divisors, all the $\alpha + g(\alpha)$ are equivalent.

If you call $R,S,T,U$ the intersection of the tangents $(PA),(PB),(AQ),(BQ)$ on your picture, this says that $(P,R) + (A,R) \equiv (P,S) + (B,S)$ and 
$(P,R) + (P,S) \equiv (A,R) + (A,T) \equiv (B,S) + (B,U)$ are all independant of $P$
If you pick a point on $E$ (for the sake of concreteness, $O = ((0,1),(\sqrt 3,1))$), you can then have an isomorphism $E \to Div^n(E)$ by $P \mapsto (n-1)O + P$ and this turns $E$ into an algebraic abelian group.
Now under this identification, for any $P$, $P + f(P) = f(O)$ and $g + g(P) = g(O)$, so
$(A,T) - (B,S) = ((A,T)+(A,R)) - ((A,R)+(P,R)) + ((P,R)+(P,S)) - ((P,S)+(B,S)) = 2g(O)-2f(O)$.
Now suppose we have a conic going through $Q$ that have share the same $4$ tangents. Furthermore, assume that it is an ellipse encompassing the circle. On our real picture, we can choose one of the two isomorphisms from $E$ to $E'$ by deciding that if $(P,Q) \in E$ then we map it to $(R,Q) \in E'$ where $R \in (P,Q]$ and not on the other side of $Q$ (this description of course breaks down when you look at the picture over $\Bbb C$, or when it is a hyperbola, or when it is an inner ellipse, so be very careful with it).
Since $f$ is independant of the choice of conic, we have $(B,S) + (Q,S) \equiv (A,T) + (Q,T) \equiv f(O)$ (where we use the isomorphism to bring everything back to $E$), and so we get $(Q,T) + (Q,S) = 3f(O)-2g(O)$ which is a constant independant of $P$.
But of course, this has to correspond to the involution $g'$ of $E'$, and so the locus of $Q$ is that conic.

Given such a conic, composing $g$ and $f$ together give you two possible translations on $E_0$ (because there are two ways to bridge $E$ with $E_0$).
Since there are as many translations on $E_0$ as points on $E_0$, this set of translations should be a twofold branched cover of a set of suitable conics.
If you start with a generic equation $ax^2+bxy+cy^2+dx+ey+f = 0$ and check when it is tangent to the lines $1 \pm i x \pm \sqrt 2 y = 0$, you get four equations
$b^2+2d^2+4cf = 4ac+8af+e^2 ; bd = 2ae ; de = 2bf ; be = 2cd$.
After ignoring a large family of degenerate equations (the family of double lines), you get $b=d=e=0$, and so $f = ac/(c-2a)$. Then you obtain the one-parameter family of conics
$(a(c-2a))x^2 + (c(c-2a))y^2 + ac = 0$.
Since this is a projective line, we should have $4$ branch points to make an elliptic curve. A branch point would be a conic where every tangent from the circle intersects the conic on a double point. Here they are :  
at $a=0$ and $c=0$ they are double lines. Every tangent from the circle has a double intersection point there (possibly at infinity). The corresponding translation map is obtained by applying the symmetry with respect to that line and switching the direction of the tangent vectors.
at $c/a=2$ you get an infinitely large ellipse. Every tangent intersects it at infinity twice, and from that direction at infinity the other tangent line is the tangent that is parallel to the first tangent, so this is just central symmetry through $(0,0)$.
at $c=a$ the conic is the circle, every tangent intersects it twice at the tangent point itself. This gives you the identity map.
So the set of translations is the twofold branched cover of the projective line ramified at $0,1,2,\infty$, and so this gives you yet another model for our elliptic curve, $k^2 = l(l-1)(l-2)$ where $l=c/a$

To finish this picture, through one point $(x,y)$ in the plane, you usually get two conics going through it. 
(because the equation $(a(c-2a))x^2 + (c(c-2a))y^2 + ac = 0$ has degree $2$ in $a,c$) 
Since the coefficients of $a^2$ and $c^2$ have distinct sign, there is one solution with $a/c < 0$ (a hyperbola), and one solution with $a/c > 0$ (an ellipse)
If you choose the hyperbola instead, you can't make the pick of isomorphism like I described and the direction of one of the tangent vectors that you think are getting related is in fact reversed compared to what you are seeing, that's why its involution will not correspond to $(Q,T)+(Q,S) = $ constant, but instead to $(Q,T) + (Q,S') = $ constant where $(Q,S')$ really doesn't have any geometrical interpretation in the original picture.

Some more geometric interpretation of the common tangents :
Ellipses are generally a bit hard to work with, however one of their points of interests are their foci, which are the real points through which the lines tangent to the ellipse have slope $\pm i$.
In our case, the tangents have slope $\pm i/ \sqrt 2$. Which means that if you scale the $y$-axis with a factor of $\sqrt 2$ (which is real, but it could be complex if the outer ellipse didn't have all the symmetries in common with the circle), you obtain a picture with three ellipses having the same tangents of slope $\pm i$, which means that they all share the same foci.
