Thursday, May 15, 2008

Hyperbolae Unveiled...Sorta

Alright guys, no shenanigans this time, I'm tired, just got off work, and would like to sleep. So I apologize if I'm a little bit frank, and a little less funny an animated in my scribe post compared to my usual ones. Hopefully the quality is still there though, if not, I'll fix it up tomorrow or w/e. Alrighty, here we go.

*note* links added under appropriate images for full sized goodness. Like I mean, it wont matter on the smartboard, since that things huge, but at home I doubt everyone has like 70 inch moniters, so ja. Thats why there there. Why arent they hotlinked or w/e its called? Cause its 1:24am haha.

Morning Class

So this mornings class wasn't much of anything special, which is kinda different from the norm (kinda paradoxical I know.) Anyways we started going through Mr. K's slides, and we solved the little equation for one of the friendly Ellipses which can be seen on the following slide.

As always, fully more large sized versions of these slides can be found in the slides themselves. Which Mr.K posts quite consistently :P

Moving on. After that we filled out a health survey. That basically took up the rest of the morning class. However, we did have a moment to do a quiz, which was marked the following afternoon class. If you were not there *cough cough* then talk to Mr.K, see if you can write it. The solutions are on the slides. If you were here, then you don't really need an explanation right? Cool.

Moving on.

Afternoon Class

The afternoon class was slightly more work oriented, and was not interrupted by some unnecessary forms/paperwork.

We began by correcting the quiz thinger from the first period.

Some usefull things to take out of that include the following.

How to distinguish Various conics from their equations.

-If the Equation has an x2 term but no y2 term it is a Horizontal Parabola
-If the Equation has an y2 term but no x2 term it is a Vertical Parabola
-If the Equation has an y2 AND an x2 term, with coefficients that are THE SAME, it is a Circle.
-If the Equation has an y2 AND an x2 term, with coefficients that are DIFFERENT, it is an Ellipse.

Remember those :)

Next we took out our previous nights homework (that being measuring, and finding the difference, of the lengths from one point on a branch of the hyperbola, to both foci. That will make more sense when the image comes up in a moment.)

The results of taking up that homework are shown as follows.

Okay, I'm going to try to explain this as best as I can, but like I said before, I just got off work, I'm crazy tired, not running on anywhere near enough sleep and a host of other not so great things. So if I phail (yes thats a "ph" fail. That doesn't mean an acidic fail but rather, means a really bad fail for all you en zero zero be's out there.)

So, the homework was to pick a point on one of the branches of the hyperbola, and measure that distance from the point, to both foci (that is to say, measure from say, point P, to F1, and then from point P to F2). We then too those values, and found the difference, (PF1-PF2). Voila! homework done. By now it should have come as no surprise that the result were values which should have been basically the same.

Once we did that, things started to get tricky. The next thing we did, was find the Vertices (sp?) of the hyperbola. To do this, you take your ruler, and line it up so it goes through both focus points. The POINTS at which the line made by the ruler intersects with the curve of the hyperbola make up it's vertices.

Okay, so if you followed my last paragraph's directions correctly, you should end up with two points at what are the vertices of the hyperbola. The next step is to connect those two points with your handy dandy ultra straight ruler. This creates what we called the TRANSVERSE AXIS. This axis can be described as the line which connects the two vertices of a hyperbola, and is seen AND labeled in green on the slide crop up there. Got it? I hope so. Moving along.

Next, we bisected this transverse axis.

"How do we do that Mr.k?"
"Well I'll show you."

Thats a rough translation of what probably everyone was thinking. Except the Ogre 1 and 2 of our class. They knew the answer already, (Outside joke.)

So to bisect the transverse axis, we folded our papers for one more time. This time, we folded one focus point, directly on top of the other. Thus we obtained a crease which defined not only the bisector, but the perpendicular Bisector of the Transverse Axis. We see this as a vertical red line on the image above. Lets step things up a notch (I know Mr. K did.)

So, do you remember the Transverse axis? and how we just cut it in two? Were going to take this line, and modify it a little bit. Using our rulers again, we measured the line from one of the FOCI (in this case F2) to that vertical line (which we now call the CONJUGATE AXIS.) This gave us some length.

Now this next bit is tricky, we took the length we measured from the PREVIOUS STEP (that being from the conjugate axis, to Focal point F2) and moved it so that one end of it was on the (whatever the singular of vertices is. Verticii?) of the nearest branch, and so that the other end of the length lies on the Conjugate axis. Since I bet your saying "what the frig" I made a quick little paint mockup to get the point across, or rather, to help you "see" what I mean.

Okay, so as I wrote on the picture, the light green is the length Conjugate Axis to F2. Then by moving it like I mentioned before, we get the makings of a triangle.

Step 3. Measure the distance from the point where the light green line meets the conjugate axis, to where the dark green axis meets the conjugate axis. This we can call length b for explanations sake. The next step is to take this length, and go from the intersection of the Transverse axis and the conjugate axis, (aka the cross in the centre) and "b" units down. I'll show you what I mean again.

So, as I wrote again, the light blue, and dark blue lines are the ones I was talking about before, and they are of equal length. Thus we get those points G1 and G2. This brings us to our next step.

By drawing perfectly 90 degree lines from the vertices (going through them, vertically) and from the end points of the light and dark blue lines (going through points G1 and G2, horizontally) we can construct a box of sorts. This is seen once again on the slides.

So you see the box right? The black dotted lines there are actually it :P So yeah after that we took our ruler again, and drew that orange cross that goes far off the page kinda. We drew that cross by going from one corner of the box to the other (and beyond.) After we had done this Mr.K revealed to us that those orange lines, are actually asymptotes for the graph. In other words, neither branch of the hyperbola will actually ever touch those orange crossey lines.

The final slide here, is the equation of the hyperbola, along with some similarities, and differences, which we came up with together, to help us remember the equation, and keep it distinct from the others.

To Conclude: The basic parts within the anatomy of a hyperbola
- Transverse Axis.
- Conjugate Axis
- Focus Points
- The Asymptotes
- And finally, you should know the equation, which is shown in the above slide image.

Alrighty, I think that about wraps this blog post up. I'm not entirely sure, because at 1:00am I kinda start to get a bit loopy and my thoughts aren't so coherent (thats probably evident based on how my writing progressed.) If you guys have any questions or whatever lemme know. I'll try to answer them, and post em up. Or if you find any errors, or know something I dont/forgot to add, lemme know so I can add it.

Well I'm off, but of course, not without letting you know who the next scribe poster is :)

Remember, vote for Zeph for VP.

K? Kthxbai

Justus- FINALLY out.

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