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.
- Since Pascal's theorem works for any conic, then the same construction will also work for a hyperbola or parabola.
- (For interest) This construction only uses a ruler, and not a compass. Therefore it is an example of a construction that works in projective geometry (i.e. the construction is preserved by any projective transformation). To make it purely projective then we should define the ellipse as the conic passing through five given points, rather than using the focal points (which are not preserved by projective transformations).
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 F′1 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 F′1 are collinear.
Since F′1 is the reflection of F1 across the line PX, then |F1P|=|F′1P|. We also know that |F1X|=|F′1X|, and so |F′1F2|=|F′1X|+|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 ΔPF2F′1 has three known sidelengths PF2 (given in the problem), F2F′1 (computed above) and PF′1=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 F′1. Therefore we can construct X as the intersection of F′1F2 with the ellipse.
Note that there are two choices of triangle ΔPF2F′1 (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 ΔPF2F′1, 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.