# First Kepler's law: The law of ellipses

**Kepler’s 1**

*Orbits of all the planets are ellipses with the Sun at one focus*^{st}law arises from the 2

^{nd}and 3

^{rd}Kepler’s laws and from the fact that the gravitational force diminishes as R

^{-2}(the inverse square of the distance law of gravity).

## Presumed knowledge

**1. Gardener**

**’s Ellipse**The sum of the distances from any point of the ellipse to its two foci is constant and equal to the major axis. This property is the essence of the construction method called the Gardener’s Ellipse:

*Keeping the rope tightened the pointed stick draws an ellipse with the two foci in the sticks (thrust into the ground) and the major axis of the length of the rope.*https://tube.geogebra.org/m/19212 (Daniel Mentrand):

**2. Reflection by an ellipse**Light rays coming from one focus ofan ellipse are reflected to the second one (All the light rays starting at one focus will be focused to a point at the other focus.)

**Proving the equivalence of those properties**The first task that was solved in Feynman's lecture was the proof of equivalence of the mentioned properties. In other words, he was to prove that

**the property of reflection (from one focus to the other) is equivalent to the property that the sum |F**

_{1}P| + |F_{2}P| is constant for any point*P*on the ellipse.Instead of proving it directly, we assign two exercises of the same nature.

**Exercise 1:**

*Use the dynamic figure below. a) Sort by size the following lengths: |F1P|+|PF2|, |F1R|+|RF_2|,*

*|F1P|+|PG|*,*|F1R|+|RG|*. b) Move the line p to align point R with point P of the ellipse. Make a conjecture about the relationship between the line p and the ellipse in this configuration.**Exercise 2:**

*An ellipse is given by its two foci and by the length of its major axis. Without drawing the ellipse, construct its movable point and the tangent of the ellipse passing through this point.*

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