# Example to show y = 2x^2 + nx + 3x - 1is a hyperbola for a specific value of n.

I will use the standard way given in many precalculus and calculus books to find a rotation of the axes so that the xy term disappears. In other words, I will follow the standard way to "get rid of the xy term" in a conic relation by trying to find x' and y' axes that are rotated with respect to the x and y axes.*

In particular, to proceed, consider the conic relation Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0.

Let x = x'cosq - y'sinq
and let y = x'sin
q + y'cosq.

Also let tan2q = B/(A - C). (This equation is a standard way to find the rotation angle, q. For more details, see the reference below or almost any calculus or honors precalculus textbook.)

With our equation, 2x^2 + nxy + 3x - y - 1 = 0, C is always 0; A = 2; D = 3; E and F both equal -1; and B = n.

Therefore, tan2q = B/A = n/2.

To view an example of how this process works, let B = n = 2sqrt(3). Then tan2q = 2sqrt(3)/2 = sqrt(3) and so 2q = p/3. Therefore, q = p/6; cosq = sqrt(3)/2; and sinq = 1/2.

Now, that means we will replace x with x'sqrt(3)/2 - y'/2
and y with x'/2 + y'sqrt(3)/2.

The result is:

2(x'sqrt(3)/2 - y'/2)^2 + 2sqrt(3)(x'sqrt(3)/2 - y'/2)(x'/2 + y'sqrt(3)/2) + 3(x'sqrt(3)/2 + y'/2) - (x'/2 + y'sqrt(3)/2) - 1 = 0

3x'^2/2 - sqrt(3)x'y' + y'^2/2 + 2sqrt(3)(sqrt(3)x'^2/4 + x'y'/2 - sqrt(3)y'^2/4) + 3sqrt(3)x'/2 + 3y'/2 - x'/2 - y'sqrt(3)/2 - 1 = 0

3x'^2/2 - sqrt(3)x'y' + y'^2/2 + 3x'^2/2 + sqrt(3)x'y' - 3y'^2/2 + 3sqrt(3)x'/2 + 3y'/2 - 1 = 0

Notice that the x'y' terms will disappear, leaving

3x'^2 - 1y'^2 + (3sqrt(3)/2 - 1/2)x' + (3/2 - sqrt(3)/2)y' - 1 = 0

And indeed, this equation is a hyperbola because the coefficients of x^2 and y^2 are both non-zero and are opposite in sign. (Of course, this sort of criterion can only be used when there is no xy term!)

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*I referenced the following book in this analysis:

Stein, S. K. (1987). Calculus and Analytic Geometry. New York: McGraw-Hill Book Company.