Conic Explorations

By: Sydney Roberts


Exploring the graph of x2+xy+y2=9 for different coefficients of the xy term.

The equation for a circle that is centered at the point (0,0) is x2+y2=r2 where r is the radius of the circle. Therefore, the equation x2+y2=9 gives us a circle centered at (0,0) with a radius of 3.


Now consider the graph x2+xy+y2=9. What shape do you think this equation makes?

We could say that this equation appears to give us an ellipse with the line y = -x being the major axis.

But what if a coefficient existed in front of the xy term? Then what would happen? LetŐs explore with some different examples    




Well it is pretty clear that the general equation x2 + axy + y2 = 9 will not yield an ellipse for all values of a. So what values of a do give us an ellipse? What other shapes will different values of a give us?

Well first, letŐs explore what happens if a is negative.


Well we see that the negative values give us the same shapes, except in a different direction. So maybe it is only the absolute value of a that determines the shape of these equations. Using the graph above, it appears that maybe when a = 2, the shape of our graph is changing. Well letŐs explore this equation some. Note that x2 + 2xy + y2 = 9 can be factored.







Which means that if a = 2, then we get two linear functions (i.e. two lines).


However, if a ­ 2, then what happens? Recall that the standard formula for a conic is       

These conics are ellipses, hyperbolas, or parabolas. To determine which shape the equation would yield, we would need to look at the discriminant: B2 – 4AC.

                      á     If the discriminant is less than 0, the function will form an ellipse.

                      á     If the discriminant is greater than 0, then the function will form a hyperbola.

Well for all of our equations that we are exploring,

A = C = 1, B varies, D=E=0, and F = -9. Hence, the discriminant  will be B2 – 4 for all of our equations (depending on B).

Hence, if |B| < 2, the equation will form an ellipse, and if |B| > 2, then the equation will form a hyperbola.