## Euclidean Construction of Tangent to Ellipse

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The applet illustrates the *Euclidean construction* of a tangent to an ellipse at a given point.

Let the point be E. Add four points on the ellipse: A, B, C, D, and consider the pentagon A, B, C, D, E. Thinking of E as a double point where two vertices of a hexagon have coalesced, we obtain a particular case of Pascal's theorem that the three intersections (AB∩DE, AE∩CD and the intersection of BC with the tangent at E are collinear.

Using this, first find P = AB∩DE and Q = AE∩CD which define a line PQ. PQ is crossed by BC at, say, R. The line ER is then the tangent to the ellipse at E.

As we see, the only tool needed to perform the job is the ruler. This makes the construction entirely projective, meaning that in the same manner one can find tangents to other conic sections, parabola and hyperbola.

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