For an algebra 2 project, I am supposed to create a drawing on a TI-84 calculator using a set of different functions (ie quadratic, absolute value, root, rational, exponential, logarithm, trigonometric and conic), but I am confused about how one would make an equation for a rotated ellipse. I know the original ellipse equation is (x^2/a^2)+(y^2/b^2)=1, and in order to graph on a calculator, you solve for y...but how do I incorporate the rotation of the axis, and then how do I solve for y so I may enter it on my calculator?
-
Here's a tip...you'll need an xy term...and even then putting it in your calculator will be really difficult!
Conic sections can all be rotated so that they are of the form
Ax^2 + Cy^2 + Dx + Ey + F = 0
All of these conics will be "lined up" to the x and/or y axis.
Parabolas only open north (C<0, A=0), south (C>0, A=0), east (C=0, A<0), or west (C=0, A>0).
Ellipses have AC>0, and have their major axis along the axis with smaller coefficient.
Hyperbolas have AC<0, and open >< when A>0, and open \/ and /\ when C>0.
To rotate parabolas, ellipses, and hyperbolas, you need to add in a Bxy term. You get a pseduodiscriminant B^2 - 4AC that tells you which conic you get. And I'm sure there's some weird formula that gives you the "angle" of this rotation, but it's a little tedious to figure out.
And as far as "solving for y", it's tough but not impossible. Three things can help you with this:
1) Pretend that the variable 'x' is really a parameter....that is, gather all terms with y^2, all terms with y (including the xy term), and all constants together.
2) Use the quadratic equation to "solve for y".
3) You'll have TWO equations, because an ellipse is never, ever, ever a function! (Can you prove this? It's a quick one-sentence proof.)
Conic sections can all be rotated so that they are of the form
Ax^2 + Cy^2 + Dx + Ey + F = 0
All of these conics will be "lined up" to the x and/or y axis.
Parabolas only open north (C<0, A=0), south (C>0, A=0), east (C=0, A<0), or west (C=0, A>0).
Ellipses have AC>0, and have their major axis along the axis with smaller coefficient.
Hyperbolas have AC<0, and open >< when A>0, and open \/ and /\ when C>0.
To rotate parabolas, ellipses, and hyperbolas, you need to add in a Bxy term. You get a pseduodiscriminant B^2 - 4AC that tells you which conic you get. And I'm sure there's some weird formula that gives you the "angle" of this rotation, but it's a little tedious to figure out.
And as far as "solving for y", it's tough but not impossible. Three things can help you with this:
1) Pretend that the variable 'x' is really a parameter....that is, gather all terms with y^2, all terms with y (including the xy term), and all constants together.
2) Use the quadratic equation to "solve for y".
3) You'll have TWO equations, because an ellipse is never, ever, ever a function! (Can you prove this? It's a quick one-sentence proof.)
-
For the calculation by hand ,you need rotation matrix(see here):
http://en.wikipedia.org/wiki/Rotation_ma…
http://en.wikipedia.org/wiki/Rotation_ma…