## Wednesday, May 14, 2008

### Hyperbolas

Today's class we didn't do a whole lot of anything other then hyperbolas. We recapped on the equation of an ellipse, refer to last class: Richard's scribe post.

We learned about how to put subscripts and super scripts, that are used to right those tiny numbers, or letters, that are used to show various things, such as an exponent on a power
Example: 21 - The 1 in this equation would be the super scripted number.
Use: This is given with <.sup> and end with <./sup> Get rid of the periods.I put them in so it wouldn't actually work.
Example2: log2 - The 2 in this log would be the subscripted number.
Use: This is given with <.sub> and end with <./sub> Get rid of the periods. I put them in so it wouldn't actually work.
This is a pretty useful html code to use. Remember you can only use these codes in the blog if you are in the "Edit Html" tab, shown beside the "Compose" tab at the top right, just above the various posting tools (spell check, insert picture, etc.).

We were given 1 choice of 3 different questions that we were asked to solve. We started by evaluating this equation: 16x2 + 9y2 = 144

The first part of the question was to change it to standard form, which was quite easy. To do this, you would have to reduce the coefficients so there wouldn't be any on the x or y values. This could be done by multiplying each side by (1/(16)9). By doing this, it would reduce the coefficients, giving us the standard form of the equation: (x2)/9 + (y2)/16 = 1.
For the 2nd part of the question, we were asked to find the centre, major axis, minor axis, vertices and the focii points.

Centre:
The centre being (h,k) but because there is no h and k values in this equation, the centre is at (0,0).

Major Axis:
the major axis being 8, because 16 = b2, and b is the length of the semi-major axis from the centre to one vertices, so square root 16, to get a semi-major length, which is 4, then multiply that by 2, to get the other semi-major axis, which gives us the major axis(two semi-majors = 1 major), which is 8.

MinorAxis:
Same idea with the minor axis, but using the a2 value, which is 9, square rooted giving us 3, which is the semi-minor axis, then multiplying by 2, to get the minor axis, which is 6. the vertices would be the endpoint of the semi-minor axes, and semi-major axes, starting at the centre (0,0).

Vertices:
The vertices of the minor-axis spanning a value of 6, from the centre and outwards horizontally, because it's a vertical ellipse, due to the fact that b2 is below the y-value (need more info?, refer to Richard's post). So if a value of 6 was spread outward horizontally, from the centre (0,0), it would give the vertices a value of (3,0) and (-3,0). The other vertices would come from the value, of 8, spanning outward vertically from the centre (0,0) giving the other vertices points at (0,4) and (0,-4).

The 3rd part of the question was to graph it: (I'll edit this in later, when the slides come out, so you can see the actual graph)

Hyperbola:
Finally, is the hyperbola, the last section of todays class, we made these hyperbola by folding a piece of paper. It started off by having a circle that was off centre on the paper, and the circle had a given centre point. We were asked to make a point about 2 cm off, out of the circle. If you were to draw a line perpendicular to the edge of the paper in landscape view that bisected this centre point, the point out of the circle should be close to touching this line. We were then asked to draw atleast 25 dots with 5 of the dots bunched in the area on the edge of the circle closest to the outside point. These points should be touching the edge of the circle. We then folded all these points, onto the points outside of the circle. This gave us an outline of a hyperbola, with the centre point of the circle, and the outside point the focus points of the hyperbola. Voila a hyperbola.

For homework, we were asked to pick a point on the left branch of the hyperbola and 2 points on the right branch of the hyperbola, after that, we measure the distance from this point to both of the focal points, then find the difference. What do you get? Find out.

That was pretty much everything we did for today's pre-cal class, and now I feel like a world-class math shaped origami teacher. I''m sure everyone feels this way, after we folded 3 different different shapes. I can't complain though, I found it quite fun. I propose the next scribe will be: eeeny meeny miineyy, Justus. Surprised?.. I thought so.

Until next time,
Francis

#### 1 comment:

m@rk said...

Francis,

I know that this comment is about a day late but they always say that its better late than never. I really like how you illustrate the step by step process of how you get the answer. It kinda shows the missing pieces of information that are not on the slides. It helps someone like me who doesn't remember a single thing from the conics section to recall everything. Thank you and job well done. Since my AP class is over , shall i say that I'll see you in class tomorrow.

-m@rk