Fun with ellipses

Before we start, clear the canvas first.

1. Setting up

Start with the focal definition of ellipse. The focal points are:

• Let

1.1 Constructing the string

Join with the focal points:

• join and
• join and

1.2 Calculating the distances

• Distance
• Distance
• Total length

Now comes the fun part. Tweak the values and so that their sum remains the same.

Draw circles with focal points as centers and as radii:

• circle
• circle

Get the points of intersection of the two circles:

• get points of intersection and highlight them

Now let's change delta:

• make to be an array
• Hide the circles around to view the points clearly
• Hide the circles around to view the points clearly

Note that these points of intersections lie on the ellipse which can be drawn like so:

• draw ellipse from focal points and

Let's reset :

• hide circle
• hide circle

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2. Some vector calculus

We can calculate the gradient of wrt the point using these steps:

• register as a param
• backpropagate the scalar output

Now the gradient of appears as an arrow at the point. Note that it is perpendicular to the ellipse.

2.1 Vector field

Once can in fact calculate the gradient of wrt each point on the grid. The \vector-field input output command shows the gradients at each integer grid:

• show vector field

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3. Reflection property

Remove the vector field and focus on the ellipse again.

• Draw a circle centered at b with radius
• Calculate the intersection of the line with the circle

If you are wondering how to figure out the variable names like line1, you can simply hover the mouse over a line or a circle to see its name.

One of the points has name p29. Connect it to p with a line:

• line from p to p29

Consider the two lines - and . It can be shown easily that these two lines have the same length (both of them added with sum to ).

The angle bisector of then reflects the line which is the same as to line :

• bisect
• highlight the bisector.

This bisector reflects the line to line . Note that it is perpendicular to the gradient at point and is tangent to the ellipse at the same point.

3.1 Proof of reflection property

First reduce the clutter by hiding unwanted figures:

• hide the circle, origin and ellipse
• zero out the gradients

The proof is based on triangle property. Pick any point on the bisector. For convenience we pick the mid-point.

• mid-point of bisector
• Create a triangle
• Hide the triangle to show the lines

By triangle inequality, we have the sum of these two sides to always be greater than or equal to this side, which equals .

Now let's draw line from to . Note that these two sides are equal.

Thus the sum of these two lines is greater than . Hence the point must lie outside the ellipse.

Every point on the bisector is outside the ellipse except point which touches the ellipse. Hence the bisector is tangent to the ellipse.

• Show the ellipse again