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Monday, April 10, 2006

Monday's Scribe

Alright, so Zaenab picked me to be the scribe for today's class. So, it was two periods today and we started the class with a problem on the board. Given x² + 121y² - 726y + 968 = 0, we were suppose to find:

-the centre and focii
-the lengths of the major & minor axes and;
-sketch the graph.

So, we had to get it into the proper form and rearrange it.

x² + 121y² - 726y + 968 = 0
x² + 121y² - 726y = - 968
x² + 121 (y² - 6y) = - 968
x² + 121 (y² - 6y + 9) = - 968 + 1089
x²/121 + 121 (y - 3)²/121 = 121/121

x²/121 + (y - 3)² = 1

a² = 121 a = 11
b² = 1 b = 1

c² = a² - b²
c² = 121 - 1
c² = 120
c = root 120 [which can be reduced to root 4 root 30, which is just 2 root 30]

centre: (0,3)
major axis: 22
minor axis: 2
focii: (±2 root 30, 3)

This is what it would look like:





A fairly understandable question, am I right? Well, we shouldn't really be having much difficulty with this, because if we are, then we're going to need some extra study time before the test. Oh yes, and because of spirit week, our test is delayed until next week. Monday, right you guys? After that question, Mr. K gave us three points for a circle:

J (-3,2)
K (4,1)
L (6,5)


With these points, we were suppose to find the centre of this circle. Discussed with the whole class, we figured out that by bisecting the line segments we could find the center. We can use systems of linear equations and use the process of elimination OR we can find the slopes and midpoints of each line segment. Yeah, I know this isn't very clear, but that's basically what I caught. We ended the period with more paper folding, horray!

Second period, we finished up the paper folding and found that we formed a hyperbola, by making a point outside of the circle but close to the edge, and with 30 individual points on the circle. We marked points P, Q, and R, as well as the focii and verticies. We learned that hyperbolas have a transverse axis and a conjugate axis.

[I couldn't upload my image]

So, the equation for a hyperbola is (x-h)² / a² - (y-k)² / b² = 1.

It is similar to the equation of an ellipse, where an ellipse is the sum of the terms, and the hyperbola is the difference. The hyperbolas open horizontally when x is positive. When y is positive, the hyperbolas open on the y-axis.

Then, Mr K. had to leave class early because he had to pick up his son. Hope he feels better Mr. K. We had about 10-20 minutes of a teacherless class, but were assigned homework. Our homework starts on page 159, questions #3 - 48, every third question.

....Lerwyne! You're the next contestant on the Price is Right! Oh, whoops, didn't mean to get your hopes up, I mean you're the next scribe. =)




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