Assignment 1: Explorations with Distance Equations

By Nikhat Parveen, UGA

In this assignment we will explore the distance equations.

Given  a pair of coordinates (x1,  y1) and (x2,  y2),the distance from any point (x, y) is given as

The distance from (x, y) to (x1,y1) is then the radius of circle where (x, y) is a point on the circle.

Consider two points (3,4) and (-5,-2). For any point (x, y) we can write the distance equations for these as

Let's see how the graph looks like for the above two equations:

Obviously circles.

Let's set each of these equations to a non constant value and see how the graph changes its shape

The following graphs represent the different distance equations and as we can see the size of the circle increase with the increase in distance that is the radius of the circle.

Let's add these two distances and see how the graph changes.

As you notice the graph, the circle has changed to an ellipse.

Now what do you notice? As the distance increases the ellipse gets bigger.

We notice from the above graphs that as the sum of distances of two equations increases the ellipse gets bigger and at a certain point changes to a circle.

Let's consider the product of the above two distance equations and see the changes in graph:

As compared to the sum the product gives a different result. The graphs for the products of two distance equation gives a circle.

We notice that with the increase in the distance the circle seems to slightly change the shape.

For different values of d......

The circle started changing shape when distance d was equal to 20 and with the increase in values if you notice from between d = 21 to 25 we see that the circles changed to oval shapes. The graph above is also called cassini ovals. The cassini ovals are quartic curves also called cassini ellipses. The  graph in red is known as lemniscates of Bernoulli and belong to a group of Cassini ovals.  For history of Cassini Ovals, please click HERE.

The distance equations gives us different results for different operations of addition and multiplication.

d. If the two given points are (-a, 0) and (a,0) then the lemniscates has its center at the origin (0,0) and major axis along the x-axis. For example, let a = 0. Then

will be this lemniscates:

The lemniscate's, also called the lemniscates of Bernoulli, is a polar curve whose most common form is the locus of points the product of whose distances from two fixed points (called the foci) a distance 2 alpha away is the constant alpha square. This gives the Cartesian equation

Simplifying the equation we get :

Graphing the lemniscates equation for different values of a, such as a = 1 to 5 gives the following graph:

Looking at the graph we can see that the lemniscates gets bigger in size and shape with increase in the value of a, ie., as the foci, the distance from the two fixed points increases the lemniscates gets bigger.

The equation   can further be simplified as:

What we did here is just simplified the given equation into a simplified form, therefore its graph remains the same.

Let's graph the equation (x2 +y2)2 = 2a2(x2-y2) + b for different values of a and b

Case i: a varying and b constant

Case ii: a constant and b varying

Case iii: a and b varying

varying a

We notice that with increase in the values of 'a' the circle which is oval in shape moves farther from the center in the x-direction but gets closer to the center in the y-direction.

varying b

Varying 'b' keeping 'a' constant, the graph moves away from the center both in the x and y direction.

varying a and b

Varying both 'a' and 'b' we can see that the graphs move away from the center in the x direction but not as much in the y direction, this is because varying 'b' tries to pull the graph away from the center as seen in the previous graph but varying 'a' the graph is pulled towards the center hence this reflects in the graph shown.

Varying b with negative values shows the following result:

So what do you notice? The graph of the cassini ovals gets parted from the center and with the increase in the negative values of 'b' changes to circles.

Let's translate this equation into polar coordinates and see how it looks graphically:

using the polar coordinates and   and substituting in the lemniscates equation  (x2 + y2)2 = 2a2(x2 - y2) we get the following:

and now using the trigonometry identities    and and substituting in the above equation we get  and simplifying it further we get the polar equation of lemniscates:

Graphically it looks.........

similar to the original lemniscates equation (x2 + y2)2 = 2a2(x2 - y2).

Varying 'a' values in the polar equation we get the following graph:

We see from the graph that the lemniscates gets bigger with increase in the value of 'a', which is what we observed earlier in the part 'd' of this assignment.