## Euclidean Construction of Tangent to Ellipse

What if applet does not run? |

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.

### Conic Sections > Ellipse

- What Is Ellipse?

- Analog device simulation for drawing ellipses
- Angle Bisectors in Ellipse
- Angle Bisectors in Ellipse II
- Between Major and Minor Circles
- Brianchon in Ellipse
- Butterflies in Ellipse
- Concyclic Points of Two Ellipses with Orthogonal Axes
- Conic in Hexagon
- Conjugate Diameters in Ellipse
- Dynamic construction of ellipse and other curves
- Ellipse Between Two Circles
- Ellipse in Arbelos
- Ellipse Touching Sides of Triangle at Midpoints
- Euclidean Construction of Center of Ellipse
- Euclidean Construction of Tangent to Ellipse
- Focal Definition of Ellipse
- Focus and Directrix of Ellipse
- From Foci to a Tangent in Ellipse
- Gergonne in Ellipse
- Pascal in Ellipse
- La Hire's Theorem in Ellipse
- Maximum Perimeter Property of the Incircle
- Optical Property of Ellipse
- Parallel Chords in Ellipse
- Poncelet Porism in Ellipses
- Reflections in Ellipse
- Three Squares and Two Ellipses
- Three Tangents, Three Chords in Ellipse
- Van Schooten's Locus Problem
- Two Circles, Ellipse, and Parallel Lines

|Activities| |Contact| |Front page| |Contents| |Geometry|

Copyright © 1996-2018 Alexander Bogomolny

67585447