Constructing tangent lines to an ellipse

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Constructing the tangent at a point on the ellipse

Below is an ellipse with focal points F1 and F2, and a point P on the ellipse. The focal points and the point P are not used in the construction, but you can click and drag them to change the shape and size of the ellipse.
Given a point A on the ellipse, we want to construct the tangent at A. To do this, construct points B, C, D and E on the ellipse, and draw the lines AB, BC, CD, DE and EA. Define X to be the intersection of AB and DE and define Y to be the intersection of BC and EA.
Pascal's theorem says that the tangent at A will intersect the line CD at a point on the line XY. Equivalently, the intersection of XY and CD is on the tangent at A. Therefore, define Z to be the intersection of XY and CD. Pascal's theorem guarantees that AZ will be tangent to the ellipse at A.

Remark.

Constructing the tangent from a point outside the ellipse

Now consider a point P outside the ellipse. Suppose for now that X is a point on the ellipse such that PX is tangent to the ellipse (we will show how to construct X below). Define F1 to be the reflection of F1 across the tangent line. The optical property of the ellipse says that a straight line F1X will reflect off the tangent and pass through F2. Equivalently, the points F2, X and F1 are collinear.
Since F1 is the reflection of F1 across the line PX, then |F1P|=|F1P|. We also know that |F1X|=|F1X|, and so |F1F2|=|F1X|+|XF2|=|F1X|+|XF2|, which is independent of X (the sum of the distances from the focal points to a point on the ellipse is constant).
Now we can construct X. The triangle ΔPF2F1 has three known sidelengths PF2 (given in the problem), F2F1 (computed above) and PF1=PF1 (given in the problem). We also know the positions of P and F2. Therefore we can construct the triangle by Euclid Prop. I.22, and hence we can construct the point F1. Therefore we can construct X as the intersection of F1F2 with the ellipse.
Note that there are two choices of triangle ΔPF2F1 (one is a reflection of the other across the line PF2), and so there is also another tangent line PY.

Another construction of a tangent line from an external point (for interest)

The previous construction uses a compass to construct the triangle ΔPF2F1, and so it is not a purely projective construction since there is no notion of a circle in projective geometry (we are only allowed to use a ruler). Therefore it is natural to ask whether there is a construction which uses only a ruler and therefore works in projective geometry.
The following construction uses only a ruler, but proving that it works requires some knowledge of the pole and polar which we will not cover in class.

Given a conic through five points A, B, C, D and E, choose two points Q and R on the conic. Let U be the intersection of PR with the conic, and let T be the intersection of PQ with the conic. Define W as the intersection of UT with RQ, and Z as the intersection of RT with UQ. Then define the points X and Y as the intersection of the line WZ with the conic. The lines PX and PY are then tangent to the conic.