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In this adventure we wish to explore lemniscates. When the graph of a product forms two loops connected at a single point a lemniscate is formed. Let us first look at an example of a lemniscate so we can get an idea of the object that we are studying. Consider the equation whose graph is shown below.

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            Many students probably view form of the equation for this lemniscate as unappealing and disgusting despite its obvious geometric appearance.  We clearly see that this equation is nothing more than the product of two circles, both with radius 3, with centers (3,0) and (-3,0).  Given the radius and centers of these circles it is clear that the origin is the only point that is on both circles and therefore it is no surprise that the origin is the point of self-intersection of the graph.  Is there a more concise or “pretty” form of this equation?  Lets explore!

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(x^2 - 6x + 9 + y^2)( x^2 + 6x + 9 + y^2) = 81


x^4 – 18 x^2 +2 x^2 y^2 +18 y^2 + y^4 + 81 = 81


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         This form is much easier on the eyes, however, we can no longer see the connection to circles.  In the end we sacrifice some intuition for visual appeal.  Now we proceed explore the general case now that we have an example to guide us.


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         The general case works exactly like the example did and so we have this nice looking representation of the lemniscate with foci (a,0) and (-a,0).  Lets plot the lemniscate for different values of a.  Notice that we only need to try nonnegative values for a since our equation is symmetric in a and –a.  Let us plot for a=2,1,1/2, and 0.


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         Notice we have no green graph for the case where a=0.  Acually when a=0 our graph is just one point, the origin.  Why is that? Well, from above we have x^2 + y^2 =0 which means we must have x=y=0 since x^2 >= 0 and y^2 >= 0 for all real values of x and y and x^2 = 0 = y^2 only when x = 0 = y.  We can deduce from above that for all nonzero a the lemniscate with foci (a,0) and (-a,0) are “essentially” the same with a=0 being a singularity case.

         What would happen if we were to vertically shift the left hand side of our “pretty” equation? That is, what does the graph of

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Look like for varying values of b?  Let us graph this for b = 2 , 1 , 0 , -1 , and -2, for a=2, just to get an idea.


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         This is interesting!  As we shift vertically up we move from two connected circles to a single figure which looks as though could possibly approach a circle as b increases.  As we shift vertically down we move from two connected loops to two disconnected loops which appear to appear to be converging to two points. 

         As a last exploration, its is hard to ignore the clear relationships and resemblences to circles so it is natural to think what this equation looks like in polar coordinates.  Recall: in polar coordinates we have x = r*cos(q) and y = r*sin(q).  If r = 0 then this equation is trival, 0 = 0, so assume r is nonzero.  Also assume a is nonzero since that is the trival case where x = y = 0. Substituting for x and y we have

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((r*cos(q))^2 + (r*sin(q))^2)^2 = 2(a^2)((r*cos(q))^2 – (r*sin(q))^2)

((r^2)((cos(q))^2 + (sin(q))^2)^2 = 2(a^2)(r^2)((cos(q))^2 – (sin(q))^2)

((r^2)(1))^2 = 2(a^2)(r^2)cos(q/2)

r^4 = 2(a^2)(r^2)cos(q/2)

r^2 = 2(a^2)cos(q/2)

(r/a)^2 = 2cos(q/2).


         This is probably the most compact form for the lemniscate but this form also hides the geometry of the lemniscate.  Certainly there is much more to explore with the lemniscate; more than we can ever known.  For example, we could explore derivatives of the function with repect to x.  What about computing the integral to find area?  What about other transformations?  Perhaps these angles will be explored in the future.