is a hyperbola when n is not 0.

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'cos****q - y'sinq** , and let **y
= x'sin****q
+ 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 + **n**cosq
sinq)x'^2
+ (-4cosq
sinq + **n**(cosq)^2 - **n**(sinq)^2)x'y' + (2(sinq)^2 - **n**cosq 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:

-4cosq 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.)

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) - **n**cosq 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!

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
= **n**cos2q/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 **w****hen 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.

*remember:***close this window to return to Amy's Write-Up #2.**