An ellipse is tangent to a straight line Given a straight line $L$: $\frac{x}{A}+\frac{y}{B}=1$, where $A$, $B>0$. Find the sufficient and necessary condition under which the following conditions hold. 


*

*The quadratic curve $E$: $a\,x^2+b\,x\,y+c\,y^2+d\,x+e\,y+f=0$ is an ellipse with the center below $L$.

*$E$ is tangent to $L$ at one point $P$ in the first quadrant.

*The center of the ellipse is below $L$ and the intercepts of the ellipse to the x-axis and to the y-axis are both positive.
My question is that do there exist some conditions in 1~3 which are redundant, i.e. can we actually remove some condition(s) such that the problem still has the same solution. Any reference, suggestion, idea, or comment is welcome, thanks!
 A: Here is a partial answer.
The type of a conic section (ellipse, hyperbola, parabola...) has to do with two determinants associated with the matrix corresponding to the quadratic equation, i.e.,
$$\tag{1}A=\begin{pmatrix}a&b/2&d/2\\b/2&c&e/2\\d/2&e/2&f\end{pmatrix}$$
i.e., its own determinant and the determinant of the upper $2 \times 2$ block
$$\tag{2}B=\begin{pmatrix}a&b/2\\b/2&c\end{pmatrix}.$$
Here is a computational approach that apparently doesn't use (1) or (2).
We can express that :
$$\tag{3} a\,x^2+b\,x\,y+c\,y^2+d\,x+e\,y+f=0$$
is the equation of an ellipse as follows by considering (3) as an equation in $y$ ; $a,b,c,d,e,f$ and x being considered as parameters. As such:
$$\tag{4}c\,y^2+(bx+e)y+(ax^2+dx+f)=0 \  \ \iff \ \ y=\dfrac{1}{2c}(-bx-e\pm\sqrt{\Delta}) \ \ \text{the pair of cartesian equations for the ellipse}$$
$$\text{where} \ \Delta:=b^2x^2+2bex+e^2-4acx^2-4cdx-4cf \ \ \text{classical discriminant}$$
Among all types of conic curves, the ellipse is charactrized as being the only  one that is bounded, moreover such that, in particular, its projection on $x$-axis is a closed interval $[x_m,x_M]$. This condition is transferred into the condition on the cartesian equations in (1) to be defined in such an interval. It means that $\Delta$ is positive for values $x \in [x_m,x_M]$ and negative elswhere. As
$$\Delta=(b^2-4ac)x^2+2(be-2cd)x+(e^2-4cf),$$
this condition is equivalent to the following double condition : 


*

*its dominant coefficient $b^2-4ac$ must be $<0$ (in order to warrant negativity for large values of $x$).

*its own discriminant $\delta:=4((be-2cd)^2-(b^2-4ac)(e^2-4cf))$ is $>0$.
But these conditions shouldn't come as a surprise. In fact, we can write:
$$\tag{5}\delta=4c(-cd^2-b^2f-ae^2+bde+4acf)$$
But the parenthesis' content in (5) is $\det(A)$ !
Now, a recap' : we can conclude that the double condition  is :


*

*$c.\det(A)>0 $ and

*$b^2-4ac<0$, which is equivalent to $\det(B)>0$...
Conclusion : the computation can be by-passed if these results are considered as known (you will find them in old books...).
