Perhaps added since this question was answered, Wikipedia has good information on this problem. There is an interesting geometric construction which contrasts with the algebraic solutions offered here: Rytz's construction.
(I have been told to add information to the answer rather than just posting links. Unfortunately as my rep is less than 50 I can't make comments yet)
The setting in which I found Rytz's construction useful was in drawing the elevation of a circle in a plan oblique projection. In this case, as in the other conjugate tangent problems that arise in parallel projection, the ellipse is tangent to the midpoints of the edges of a parallelogram. This is a slightly more constrained and regular situation than the diagram referenced in the original question, though a tangent parallelogram could easily be constructed around the ellipse shown in that image.
Rytz's construction is apparently the last refinement of a long series of solutions to this problem, starting with Pappus. It relies on the fact that conjugate diameters are affine images of perpendicular diameters of a circle. In particular, the perpendiculars from the foci to any tangent intersect the tangent on the auxiliary circle, the circle centered at the centre of the ellipse with the major axis as diameter. As I understand it Rytz's construction is a carefully minimized (in terms of number of steps) derivative of the earlier techniques, intended for practical use in drafting, etc.
Any second-degree curve equation can be written as
$$Ax^2+Bxy+Cy^2+Dx+Ey+F=0\tag{1}$$ or $$ax^2+2hxy+by^2+2gx+2fy+c=0\tag{2}$$ where $$A,B,C,D,E,E,a,b,c,f,g,h\in\mathbb R$$
To find type of conic and nature of conic we use $\Delta$ and is given by $$\Delta=\begin{vmatrix}a&h&g\\h&b&f\\g&f&c\end{vmatrix}$$ $$=abc+2fgh-af^2-bg^2-ch^2\tag{3}$$
If $\Delta$ is $0$, it represents a degenerate conic section. Otherwise, it represents a non-degenerate conic section.
Also, the type of conic section that the above equation represents can be found using the discriminant of the equation, which is given by $B^2-4AC$.
Conditions regarding the quadratic discriminant are as follows:
If $\Delta=0$:
$\bullet$ If $h^2-ab\gt0$, the equation represents two distinct real lines.
$\bullet$ If $h^2-ab=0$, the equation represents parallel lines.
$\bullet$ If $h^2-ab\lt0$, the equation represents non-real lines.
If $\Delta\neq0$:
$\bullet$ If $B^2-4AC\gt0$, it represents a hyperbola and a rectangular hyperbola $(A+C=0)$.
$\bullet$ If $B^2-4AC=0$, the equation represents a parabola.
$\bullet$ If $B^2-4AC\lt0$, the equation represents a circle $(A=C,B=0$) or an ellipse $(A\neq C)$. For a real ellipse, $\Big(\frac\Delta{a+b}\lt0\Big)$.
So for the given case, the equation of conic is (after putting all the points in the general equation of conic and finding all the coefficient) $$ax^2-axy+4ay^2-4a=0$$
Comparing above equation with equation (1) and (2) we get the values of required coefficient as $A=a$, $B=-a$ and $C=4a$ so we get the value of $B^2-4AC=-3a^2$ and also from equation (3) we get $\Delta=-15a^2$. Since $\Delta\neq0$ and $B^2-4AC\lt0$ and also $A\neq C$, so as we know, this is the condition of an ellipse.
$\therefore$ Conic $ax^2-axy+4ay^2-4a=0$ is a real ellipse.
Best Answer
It depends.
Take an example.
Whatever major and minor axis are given to you, shift and rotate axes such that the major axis is new x-axis and minor axis is new y-axis. Now, if you are given points, say, $(3,0)$ and $(-3,0)$ (in new coordinate system). It may be ellipse with equation $$\frac{x^2}{9}+\frac{y^2}{k^2}=1$$ (for some k) or it may be a hyperbola with equation $$\frac{x^2}{9}-\frac{y^2}{k^2}=1$$
But, say, if your points are like $(3,0)$ and $(0,2)$, you know it is ellipse with equation, $$\frac{x^2}{9}+\frac{y^2}{4}=1$$
Hope it helps:)