Given both foci, a point on an ellipse and a random secant line how can one construct the meeting of the curves We are given three points: $F_1, F_2$ that are the two foci of the ellipse $\mathcal E$ and point $A$ that belongs to $\mathcal E$.
We are also given a random line $\ell$ (let's assume it does not contain $A$).
How can we find points $B$ and $C$ such that $\{B,C\} = \mathcal E \cap \ell \neq \emptyset$ using straight edge and compass?
We can easily get:

*

*center of $\mathcal E$ as it is the midpoint of $F_1F_2$

*axes of symetry of $\mathcal E$

*major axis:

$X = \odot(A,AF_1) \cap \overleftrightarrow{AF_2}$ (with $X$ and $F_2$ in disjointed halfplanes of $AF_1$).
if $M$ is the midpoint of $XF_2$, then $a = F_2M$ so we have the vertices of the ellipse.
 A: Construct centre and vertices of the ellipse, as explained in the question, as well as semi-axes $a$ and $b$. Let $Q$ be the point of intersection between lines $\ell$ and $F_1F_2$. Take any point $P$ on $\ell$ (different from $Q$) and its projection $H$ on $F_1F_2$. Construct then point $P'$ on ray $HP$ such that $P'H:PH=a:b$; line $\ell'=QP'$ is the image of $\ell$ under a dilation of ratio $a/b$ perpendicular to the major axis of the ellipse. (If $\ell$ is parallel to $F_1F_2$, then $\ell'$ is parallel to $\ell$ and can be constructed in the obvious way).
Construct then the auxiliary circle, centred at the centre of the ellipse and with radius $a$: this circle, too, is the image of the ellipse under a dilation of ratio $a/b$ perpendicular to the major axis of the ellipse. If $B'$ and $C'$ are the intersections of this circle with line $\ell'$, we can then easily construct $B$ and $C$ as the intersections of $\ell$ with the perpendiculars from $B'$ and $C'$ to the major axis of the ellipse.

