Epic Item 10 (1.4)
by EpicSkills32, Nov 13, 2013, 5:20 AM
Yes, I am finally going to continue. As stated before,
EpicSkills32 wrote:
Conic sections are cool. When the intersect with a plane, they can form parabolas, circles, ellipses, hyperbolas, a line, two lines, a point, etc.
The standard form of a conic section equation is :
$\displayset Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0 $
Generally, they are pretty simple to graph with a pencil, graph paper, and straightedge. However, the
term is tricky. Usually, you get an equation without it, and the graphing is pretty straightforward. However, if there is a B value that is not zero, it gets tricky. Here's what the B does: it rotates the equation. If the B is not given, there is a convenient formula for finding the angle of rotation. 
.
In my next few posts, I will be deriving the B value. Usually when graphing an equation with a B, the term is eliminated by the following equations:
.
The standard form of a conic section equation is :
$\displayset Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0 $
Generally, they are pretty simple to graph with a pencil, graph paper, and straightedge. However, the
term is tricky. Usually, you get an equation without it, and the graphing is pretty straightforward. However, if there is a B value that is not zero, it gets tricky. Here's what the B does: it rotates the equation. If the B is not given, there is a convenient formula for finding the angle of rotation. 
.In my next few posts, I will be deriving the B value. Usually when graphing an equation with a B, the
term is eliminated by the following equations:
.So here goes.
We start with the general conic equation.
$\displayset Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0 $
We're looking for the B value in terms of the rest of the equation. We are given a way to eliminate the xy term, by the following equations.
We will plug those in and see what happens.
Simplifying a bit we get:
$\ A\left( (x^\prime\cos\alpha)^2 - 2x^\prime y^\prime\cos\alpha\sin\alpha+(y^\prime\sin\alpha)^2 \right) +B\left(x^\prime ^2 \cos\alpha\sin\alpha - x^\prime y^\prime(\sin\alpha)^2 + x^\prime y^\prime(\cos\alpha)^2 - y^\prime ^2 \cos\alpha\sin\alpha\right) +C\left( (x^\prime\sin\alpha)^2 + 2x^\prime y^\prime \cos\alpha\sin\alpha + (y^\prime\cos\alpha)^2\right) + Dx^\prime \cos\alpha - Dy^\prime \sin\alpha + Ex^\prime\sin\alpha + Ey^\prime\cos\alpha + F = 0 $.
EDIT: why are the plus signs lined up like that? That's weird. XP
And that's it for today. Check back as I continue this process(and hopefully finish it).
~ES32
This post has been edited 6 times. Last edited by EpicSkills32, Oct 8, 2014, 5:40 AM
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