(part 0.5)

Before I derive the B term in the conic equation(if you don't what this is, look about 2 posts down), I'm going to present a sample solution of how you would eliminate the term.
In order to graph a conic equation by standard means, we must get rid of the rotation. We do so by finding the angle of rotation, and plotting new axes. Then, we graph the simplified equation as normal on these new axes.
Here's an exmaple:

Graph the equation: .

Analysis: It's a conic equation. There's also an xy term. We will have to get rid of that.

Solution:
We have a formula(actually 2) for getting rid of the xy term.
$\displayset x = x^\prime \cos \alpha - y^\prime \sin\alpha $.
$\displayset y = x^\prime \sin \alpha + y^\prime \cos \alpha $.

However, we have to find . To do so, we have another equation: (A, B, and C being the terms in the generic conic equation; refer to other post).

Plugging in, we get . From our knowledge of trigonometry, we know that . (and any other coterminal angles). We will use for simplicity and ease. It follows that .

Plugging into the elimination equations(and solving the cosines and sines), we get:
$\displayset x = \dfrac{x^\prime - y^\prime}{\sqrt2} $
$\displayset y = \dfrac{x^\prime + y^\prime}{\sqrt2} $

Now, all we have to do is plug these values of x and y into the original equation we're graphing.



Now we square and get rid of the denominators. This is why we left the in the denominators.

$\frac{1}{2} (x^\prime ^2 -2x^\prime y^\prime + y^\prime ^2) - (x^\prime ^2 - y^\prime ^2) + \frac{1}{2} (x^\prime ^2 + 2x^\prime y^\prime ^2 ) - 6 (x^\prime - y^\prime) - 2 (x^\prime + y^\prime) = -18 $.

Whew. Now we multiply everything by two to get rid of the fractions.

$\ x^\prime ^2 -2x^\prime y^\prime + y^\prime ^2 - x^\prime ^2 - y^\prime ^2 + x^\prime ^2 + 2x^\prime y^\prime ^2 - 12 (x^\prime - y^\prime) - 4 (x^\prime + y^\prime) = -36 $.

And then we simplify the rest of the parentheses.

$\ x^\prime ^2 -2x^\prime y^\prime + y^\prime ^2 - x^\prime ^2 - y^\prime ^2 + x^\prime ^2 + 2x^\prime y^\prime ^2 - 12x^\prime - 12y^\prime - 4x^\prime + 4y^\prime = -36 $.

Grouping like terms gives:

$\ 4y^\prime ^2 + 8 y^\prime - 16 x^\prime = -36 $.

Dividing by 4 results in:

$\ y^\prime ^2 + 2 y^\prime - 4 x^\prime = -9 $.

It's a parabola! (Again, by analytical geometry knowledge) We complete the square on the left side to put it into standard form of a parabola.

$\ y^\prime ^2 + 2y^\prime + 1 - 4x^\prime= -8 $.

.





Now we know that the translated origin for this rotated equation is at . So our final equation is:
$\ \left(y^\prime^\prime\right)^2=4\left(x^\prime^\prime\right) $

Now we start working on graphing. Let's figure out which way this parabola is facing. Since the y term is squared, we know that it is a left-right parabola. But does it face left or right? The p value is positive, so we know that it faces right.
In case you don't know this

Now since the equation is rotated by an angle of , the axes will form a symmetrical octagonal sort of thing. Basically, the 90 degree angles of the original axes will be bisected. The line that makes the positive 45 degree angle with the original axis will be the axis, and the line that makes the positive 125 degree angle with the original axis will be the axis. (It's perpendicular to the axis.)
The graph is going to have a parabola facing(opening) north-east.

We need a few key points.
Let's start out with the vertex. The vertex of a parabola is given by the following: . But since the way to graph a parabola is to find the translated origin and use that as the vertex, we already found it. It's on the rotated axes.
How about the focus. The focus of a parabola can be found by the following: . Well, the p value is 1, so the focus is 1 away from the vertex along the axis of symmetry (remember, this is positive, because the p value is positive.) In case you still can't picture it, it's at on the rotated, translated axes.
For our final points, we will find the ends of the latus rectum. The latus rectum length for several of the conics(ellipse, parabola, hyperbola, etc.) can be found by . Substituting yields: . Now we know that on both sides of the focus, we go out 2 units. The points would be .
Now we have enough to draw a decent graph.

Unfortunately, I am not very familiar with asy, so here's a picture of my graph.

In analysis, the graph is a bit too hyperbolic. (I've been graphing hyperbolas recently.) The graph of a parabola should be more focused in a general direction(obviously, this depends on eccentricity). The sides are sloping almost as if approaching asymptotes. The sides should be pointing in a general direction of north-east, not in 2 such different directions.

I hope you enjoyed and survived this long process of graphing a rotated conic.

Next time, I will be deriving the B term(as advertised below).

~ES32