# PROOF that y = 2x^2 + nx + 3x - 1is a hyperbola when n is not 0.

### Substituting and Expanding

Here is the expansion of our equation 2x^2 + nxy + 3x - y - 1 = 0 with the substitutions made for x and y.
That is, let x = x'cosq - y'sinq , and let y = x'sinq + y'cosq, and substitute:

2(x'cosq - y'sinq)^2 + n(x'cosq - y'sinq)(x'sinq + y'cosq) + 3(x'cosq - y'sinq) - (x'sinq + y'cosq) - 1 = 0

becomes

(2(cosq)^2 + ncosq sinq)x'^2 + (-4cosq sinq + n(cosq)^2 - n(sinq)^2)x'y' + (2(sinq)^2 - ncosq sinq)y'^2 +(3cosq - sinq)x' + (-3sinq - cosq)y' -1 = 0

We want the x'y' term to disappear, so we want the coefficient of x'y' to be zero:
-4cos
q sinq + n(cosq)^2 - n(sinq)^2 = 0

Therefore,
n(cos2q) = 2(sin2q), [since (cosq)^2 - (sinq)^2 = cos2q and 2cosq sinq = sin2q]

which means the x'y' term will disappear if tan2q = n/2 (= B/A, as mentioned in the example.)

### Investigating the coefficients of x^2 and y^2

So if tan2q = n/2, the x'y' term disappears. The question is whether the coefficients of x^2 and y^2 will be non-zero and opposite in sign (which will indicate that the equation represents a hyperbola.)

The coefficient for x'^2 is 2(cosq)^2 + ncosq sinq

And the coefficient for y'^2 is 2(sinq)^2 - ncosq sinq = 2(1 - (cosq)^2) - ncosq sinq = 2 - 2(cosq)^2 - ncosq sinq.

So, essentially, if the coefficient for x^2 is Q, the coefficient for y^2 is 2 - Q, which means the coefficients for x^2 and y^2 always add up to 2. Interesting!

### The problem of SIGNS...

But this "discovery" is a little problematic when considering the signs of Q and 2 - Q. That is, if Q > 0, 2 - Q is positive for Q between 0 and 2, and negative for Q > 2. (See example, where Q = 3 and 2 - Q = -1.) Therefore, potentially, both Q and 2 - Q will be positive for 0 < Q < 2.

To investigate this rather problematic development, I made a movie. In the movie, the purple graph represents the coefficient of x^2; the blue graph represents the coefficient of y^2; and the red graph is the line y = 2. To get the equations for the graphs of the coefficients, I used the fact that tan2q = n/2 means sin2q = ncos2q/2, which means sinq = n^2 cos2q/4.
So
2(cosq)^2 + ncosq sinq = 2(cosq)^2 + n^2cos2q/4.
I also used the fact that
q = 0.5arctan(n/2). Therefore, I could graph the coefficient of x^2 as

y = 2(cos(0.5arctan(n/2)))^2 + n^2 cos2(0.5arctan(n/2))/4

and the coefficient of y^2 as

y = 2 - (2(cos(0.5arctan(n/2)))^2 + n^2 cos2(0.5arctan(n/2))/4)

In the movie, n ranges from -50 to 50. View the movie now, and when you finish viewing the movie, press the Back button to return to this page.

Notice--WOW--that the coefficient for x^2 is never less than 2, and the coefficient for y^2 is never greater than 0. Therefore, the graph must always produce a hyperbola when n is not zero, because, when the xy term disappears, the coefficient for x^2 will always be positive and the coefficient for y^2 will always be negative. This result can be justified algebraically, I'm sure, but for now, we'll let the technology do the work. QED.