**Solution to Sketchpad Problem 3**

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This sketch was based on the fact that a constant distance, K, (represented by CD in the sketch below) can be divided into two parts, d1 and d2=K-d1. The distances d1 and d2 act as the distances from A and B (see the sketch) to a point (x,y) called F on an ellipse, respectively.

The resulting formula is based on the accompanying sketch, where

a=the distance from the center to the point at which the ellipse crosses the x-axis

b=the distance from the center to the point at which the ellipse crosses the y-axis

c=the distance from the center to point A or point B

Starting with a, we see that a=SQRT(b^2 + c^2).

Further, d1 + d2 = K = 2a.

But d1=SQRT((x+c)^2 + y^2) and d2=SQRT((x-c)^2 + y^2).

Then SQRT((x+c)^2 + y^2) + SQRT((x-c)^2 + y^2) = 2a.

Switching things around and squaring both sides, we obtain:

(x+c)^2 + y^2 = 4a^2 - 4a * SQRT((x-c)^2 + y^2) + (x-c)^2 + y^2
Simplifying, we get

SQRT((x-c)^2 + y^2) = a - (c/a) * x
Squaring again and simplifying:

(x/a)^2 + (y/(a^2 - c^2))^2 = 1
Finally, we obtain the equation for an ellipse,

(x/a)^2 + (y/b)^2 = 1.
By moving point E, we obtain a collection of points forming an ellipse.

If you would like to see the formation of this ellipse, click here.

Click here to return to Our Group's Homework 2.