WLOG the equations of the Circle & the Ellipse can be chosen to be $$\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1\text{ and } x^2+y^2=r^2$$
We know the parametric coordinate of any $P(x,y)$ on the ellipse is $(h+a\cos\phi,k+b\sin\phi)$ where $\phi$ is the eccentric angle
To find the intersection, $$(h+a\cos\phi)^2+(k+b\sin\phi)^2=r^2$$
Use Weierstrass substitution, to form a Bi-Quadratic Equation in $\tan\frac\phi2$ whose roots are $\displaystyle\tan\frac\alpha2,\tan\frac\beta2,\tan\frac\gamma2,\tan\frac\delta2$
Now using $\tan(A+B+C+D)$ formula ,
$$t^4\{(h-a)^2+k^2-r^2\}+(4bk)t^3+()t^2+(4bk)t+()=0$$
Using Vieta's formula $$\sum \tan\frac\alpha2=-\frac{4bk}{(h-a)^2+k^2-r^2}=\sum \tan\frac\alpha2\tan\frac\beta2\tan\frac\gamma2$$
$$\implies\tan\left(\frac{\alpha+\beta+\gamma+\delta}2\right)=0\implies \frac{\alpha+\beta+\gamma+\delta}2=n\pi$$ where $n$ is any integer
I have no idea about the parabola, I'll only cover the case of ellipse.
Since "concaveness" is preserved under linear transform, we can use a linear transform
to map the ellipse to the unit circle centered at origin. The question becomes "can we inscribe a concave quadrilateral into that unit circle?".
Let $ABCD$ be a concave simple quadrilateral. Since it is concave, one of its vertices, say $D$, lies in the interior of the triangle $ABC$.
This means we can find $\alpha, \beta, \gamma > 0$ such that
$$D = \alpha A + \beta B + \gamma C\quad\text{ and }\quad\alpha+\beta+\gamma + 1$$
Let $\lambda = \alpha+\beta$ and $\mu = \frac{\alpha}{\lambda}$, we have
$\lambda,\mu \in (0,1)$ and $(\alpha, \beta, \gamma) = (\lambda\mu,\lambda(1-\mu),1-\lambda)$
Let $\varphi(P)$ be the squared distance of point $P$ from origin. When $A,B,C$ lies on the unit circle, we have $\varphi(A) = \varphi(B) = \varphi(C) = 1$. In order to inscribe $ABCD$ into the unit circle, we also need $\varphi(D) = 1$.
As a function of $P$, $\varphi(P)$ is strictly convex over the Euclidean plane. This implies
$$\begin{align}
\varphi(D)
&= \varphi(\alpha A + \beta B + \gamma C)
= \varphi(\lambda(\mu A + (1-\mu)B) + (1-\lambda) C)\\
&< \lambda \varphi(\mu A + (1-\mu)B) + (1-\lambda) \varphi(C)\\
&< \lambda( \mu \varphi(A) + (1-\mu)\varphi(B)) + (1-\lambda) \varphi(C)\\
&= \lambda( \mu + (1-\mu) ) + 1-\lambda\\
&= 1
\end{align}
$$
As a result, $D$ falls inside the interior of unit disk and cannot lie on the unit circle.
Update
I find an argument that works for both parabola and ellipse. The key is the concept extreme point of a convex set. There are several easy to verify facts.
- If $D$ lies in the interior of triangle $ABC$, then $D$ is not an extreme point for the triangle (which is a convex set).
If $P \in X \subset Y$ for convex sets $X$ and $Y$, then $P$ is an extreme point for $Y$ implies $P$ is an extreme point for $X$.
The region bounded by an ellipse is convex and the ellipse is the set of extreme points for this region.
Let's say we have an ellipse $\mathcal{E}$ circumscribe a concave quadrilateral $ABCD$. Since $ABCD$ is concave, one of its vertices, say $D$ lies in the interior of triangle $ABC$.
- By $(1)$, $D$ is not an extreme point for triangle $ABC$.
- By $(2)$, $D$ is not an extreme point for the region bounded by $\mathcal{E}$
(since this region contains triangle $ABC$).
- By $(3)$, $D$ doesn't lie on $\mathcal{E}$.
This contradict with the assumption that $\mathcal{E}$ circumscribe $ABCD$.
For parabola, the argument is similar. A parabola divides the plane into
two components. One component of it is convex while the other concave.
It is easy to see if a parabola $\mathcal{P}$ circumscribe an quadrilateral $ABCD$, then the interior of the edges $AB$, $BC$, $CD$, $DA$ and body of quadrilateral lies inside the convex component.
We can replace $(3)$ by an alternate fact.
- The parabola is the set of extreme points for the corresponding convex component.
Essentially same argument like before tell us if $ABCD$ is concave, then the
point $D$ (the one lies inside the triangle formed by other points)
cannot lies on $\mathcal{P}$. A contradiction again!
Best Answer
Since it's lack of the contexts for the cases of ellipse and parabola, I'd like to start from a circle:
$$(x,y)=r(\cos \theta,\sin \theta)$$
Equation of chord $AB$
$$x\cos \frac{\alpha+\beta}{2}+y\sin \frac{\alpha+\beta}{2} =r\cos \frac{\alpha-\beta}{2}$$
$$m_1=-\cot \frac{\alpha+\beta}{2}$$
From properties of cyclic quadrilaterals:
$$180^{\circ}=A+C=B+D$$
$$0=\tan B+\tan D$$
$$0=\frac{m_1-m_2}{1+m_1 m_2}+\frac{m_3-m_4}{1+m_3 m_4}$$
which is equivalent to
$$0=\tan \frac{\alpha-\gamma}{2}+\tan \frac{\gamma-\alpha}{2}$$
For an ellipse $(x,y)=(a\cos \theta,b\sin \theta)$
$$m_1=-\frac{b}{a}\cot \frac{\alpha+\beta}{2}$$
Transform every $m_i$ back to the auxiliary circle,
$$\frac{a}{b}(m_1-m_2+m_3-m_4)= \frac{a^3}{b^3}(m_2 m_3 m_4-m_1 m_3 m_4+m_1 m_2 m_4-m_1 m_2 m_3)$$
For a hyperbola $(x,y)=(a\cosh t,b\sinh t)=(a\cos it,-bi\sin it)$
$$m_1=\frac{b}{a}\coth \frac{t+u}{2} =\frac{bi}{a} \cot \frac{i(t+u)}{2}$$
$\fbox{Note that $t, u, v, w$ differ from the eccentricity angles $\alpha, \beta, \gamma, \delta$ in $(a\sec , b\tan )$.}$
Now replace $b$ by $bi$, we have
$$\frac{m_1-m_2+m_3-m_4}{m_2 m_3 m_4-m_1 m_3 m_4+m_1 m_2 m_4-m_1 m_2 m_3} =\frac{a^2}{(bi)^2}$$
For a parabola of the form $x^2=4py$, take $b\to \infty$
For a parabola of the form $y^2=4qx$, take $a\to \infty$.