Any triangle can be circumscribed in a unique circle (the circumcircle) - that is, a circle passing through the three vertices.
Can any convex quadrilateral be circumscribed by an ellipse? Under what conditions are there zero / exactly one / finite multiple / infinite solutions?
Yes, you can always find at least one circumscribing ellipse, assuming strict convexity (i.e. that no three points are collinear).
Let $ABCD$ be a convex quadrilateral. The sum of the four interior angles is $360^\circ$; and if opposite angles sum to $180^\circ$, we have a cyclic quadrilateral, and we are done.
So now assume without loss of generality that $\angle ABC + \angle ADC > 180^\circ$. Also, for ease of visualisation, picture it so that $AC$ is vertical.
By the strict convexity of $ABCD$, $B$ and $D$ lie on opposite sides of $AC$. So as we stretch the plane in a horizontal direction, $\angle ABC$ and $\angle ADC$ tend continuously to zero. By the intermediate value theorem, there must exist a stretch factor $\sigma$ for which $\angle ABC + \angle ADC = 180^\circ$. Now we have a cyclic quadrilateral! So we can circumscribe the stretched $ABCD$ in a circle $S$.
Now just squeeze the plane back again by a factor of $\sigma$, so that $B$ and $D$ return to their starting positions. The circle $S$ has now become a circumscribing ellipse.
Updated to add: There is no need to stretch the plane in a horizontal direction; any direction but the vertical would do. And each direction of stretch results in a different circumscribing ellipse. So as long as the original quadrilateral is not cyclic, there is an infinite family of circumscribing ellipses.
My feeling is that there is an infinite family of circumscribing ellipses even if the original quadrilateral is cyclic. But I haven't shown that here.
Can any convex quadrilateral be circumscribed in an ellipse ?
Hint: Five points determine a conic. Depending on where you choose this fifth point to be, you can have an infinity of various conic sections $($ellipses, parabolas, hyperbolas$)$ passing through those initial four points.
Five points determine a Conic. Depending on the relative position of the fifth point with respect to vertices of a quadrilateral we get hyperbolas, ellipses and parabolas. Many Java programs like GSP, Geogebra, Z.u.L show this by means of appelets.
Suppose the fifth point is nearer to point of intersection of the diagonals of the quadrilateral you get a hyperbola.
Taking the equation of conic classically
$ a x^2 + 2 h x y + b y^2 + 2 f x + 2 g y = c $
( when divided by any coefficient only 5 points are needed to define a conic).
$ A x_1^2 + 2 H x_1 y_1 + B y_1^2 + 2 F x_1 + 2 G y_1 = 1 $
Write the conic equation as a (6 X 6) determinant and put in the condition for the equation to represent a pair of straight lines.
It defines a criterion to delineate concave/convex cases.
It defines a unique ( no finite multiple / infinite) solution.
You can do it now I think.
Yes, any convex quadrilateral can be circumscribed by an infinite number of elipses.
In the proof I draw a distinction between two cases: 1) a convex quadrilateral is not a parallelogram; 2) a convex quadrilateral is a parallelogram
Proof
Without loss of generality, let $L_1\equiv m_1x -y +r_1=0$, $L_2\equiv m_2x -y +r_2=0$, $L_3\equiv m_3x -y +r_3=0$, $L_4\equiv m_4x -y +r_4=0$ be the equations of lines $AB$, $BC$, $CD$, $DA$ of a quadrilateral ABCD.
Let us suppose ABCD is not a parallelogram. Consequently at least there is a pair of opposite sides which aren't parallel, say, AB and CD. Therefore we may assume $m_1\neq m_3$.
All the conics which circumscribe the quadrilateral ABCD can be given by the equation $kL_1L_3+L_2L_4=0$ (except the degenerate conic consisting of the pair of lines $AB$ and $CD$).
Therefore all the conics circumscribing the quadrilateral (except the mentioned degenerate conic) are given by the equation $$k(m_1x -y +r_1)(m_3x -y +r_3)+(m_2x -y +r_2)(m_4x -y +r_4)=0,$$ $$(m_1m_3k +m_2m_4)x^2-((m_1+m_3)k+(m_2+m_4))xy+(k+1)y^2+...=0$$
The type of this circumscribing conic is given by
$$\delta=(km_1m_3+m_2m_4)(k+1)-\frac 14(k(m_1+m_3)+(m_2+m_4))^2,$$
This circumscribing conic is an ellipse if $\delta>0$, an hyperbole if $\delta<0$, a parabola if $\delta=0$.
Developing $\delta$ we get
$$4\delta= 4m_1m_3k^2+4m_2m_4k+4m_1m_3k+4m_2m_4-((m_1+m_3)^2k^2+2(m_1+m_3)(m_2+m_4)k+(m_2+m_4)^2),$$ $$4\delta=-(m_1-m_3)^2k^2+2[2m_2m_4+2m_1m_3-(m_1+m_3)(m_2+m_4)]k-(m_2-m_4)^2,$$
$$4\delta=-(m_1-m_3)^2k^2+2[(m_1-m_3)(m_2-m_4)+2(m_1-m_4)(m_3-m_2)]k-(m_2-m_4)^2$$
Now let $\psi$ be the discriminant of this second degree equation in k: $$\psi=4[(m_1-m_3)(m_2-m_4)+2(m_1-m_4)(m_3-m_2)]^2-4(m_1-m_3)^2(m_2-m_4)^2,$$ $$\frac 14 \psi=[(m_1-m_3)(m_2-m_4)+2(m_1-m_4)(m_3-m_2)]^2-(m_1-m_3)^2(m_2-m_4)^2,$$ $$\frac 14 \psi=4(m_1-m_3)(m_2-m_4)(m_1-m_4)(m_3-m_2)+4(m_1-m_4)^2(m_3-m_2)^2,$$ $$\frac 1{16} \psi=(m_1-m_3)(m_2-m_4)(m_1-m_4)(m_3-m_2)+(m_1-m_4)^2(m_3-m_2)^2,$$ $$\frac 1{16} \psi=(m_1-m_4)(m_3-m_2)[(m_1-m_3)(m_2-m_4)+(m_1-m_4)(m_3-m_2)],$$ $$\frac 1{16} \psi=(m_1-m_4)(m_3-m_2)(m_1-m_2)(m_3-m_4),$$
$$\psi=16(m_1-m_2)(m_2-m_3)(m_3-m_4)(m_4-m_1)$$
As a corollary of the theorem stated and proved in The concave quadrilateral and the slopes of its sides,
ABCD is convex quadrilateral $\Rightarrow$ $\psi>0$ $\Rightarrow$ the equation $\delta=0$ admits two distinct solutions, say, $k_1$ and $k_2$ ($k_1<k_2$).
Consequently, since $-(m_1-m_3)^2<0$, for $k_1<k<k_2$ we get $\delta>0$ and the quadrilateral is circumscribed by the ellipses given by the equation $kL_1L_3+L_2L_4=0$,
QED.
If ABCD is a parallelogram, then $m_1=m_3$ and $m_2=m_4$ and
$$\delta=4(m_1-m_2)^2k$$
Consequently for $k>0$, $\delta>0$. As a result, for $k>0$ ABCD is circumscribed by ellipses given by the equation $kL_1L_3+L_2L_4=0$,
QED.
