Saturday, May 17, 2008

WORKING WITH A(N) HYPERBOLA - CONCLUSION [beware shameless advertising.]

Sorry the post is late, but I guess this is why I needed the weekend. Thank you Justus. Nice doodle there, by the way. I don’t think I can incorporate images of the slides in my post, but I’ll still try to use some form of an outline and I’ll also parody a certain [we]blogging method. Yes, I said PARODY. I thank Richard for this.

On Friday’s class, our goal was to conclude our lesson about the hyperbola. We leaed to: rn
  • Write the standard forms of a hyperbola [horizontal and vertical]
  • Find the similarities and differences between vertical and horizontal hyperbolas
  • Find the Pythagorean theorem property in hyperbolas
  • Finally, apply everything we’ve learned by graphing and finding the coordinates required in every hyperbola.
  • As well as that, we also had a sort of English epiphany, but we’ll discuss that later.


On Thursday, the standard formulas of a hyperbola were:
HORIZONTAL: [[(x – h)2 / a2] - [(y - h)2 / b2]]
VERTICAL: [[(y – h)2 / a2] - [(x - h)2 / b2]]
As soon as the class first caught a glimpse of the equations, we wondered why it was the same as the ellipse, but as we examined closely, there were differences as well as there were similarities between them.



  • they have the same terms
  • they both equal 1
  • both have the same denominators
  • both have square binomials
  • (h, k) indicates centre


  • numerators have different orientation
  • numerators have opposite signs

We also emphasized the differences between the ellipse and the hyperbola by cloning the slides from the ellipse and altered them. We rewrote the formulas and redrew the diagrams, interactively of course. There was even a point when Mr. K was going to redraw a diagram that was on the other page. For the sake of elegancy that rubbed off on the class, they argued for him to clone it, cut it, paste it and shrink it…[that’s sounded like I was singing a new version of “Technologic” or something haha DAFT PUNK man. All the way.] The point is “Just Shrink It”, man. That should be the new Nike slogan. Jeebers, I shouldn’t call out on Mr. K, it’s not even June yet. By the way, it was deemed necessary to rotate the “O” in the graph.

Anyhow, we then realized that a2 [pronunciation: “ay squared”] could be smaller than b2 [bee squared] after seeing the animations that proved so. These diagrams revealed what would happen if there were variations in the lengths of a [semi-transverse] and b [semi-conjugate]. There were also additional animations for the variation of points h [ey-ch] and k [kay]. “h” moves center and hyperbolas horizontally because it is an “x” coordinate. “k” moves center and hyperbolas vertically since it is a “y” coordinate.

The class then asked themselves this question: if a2 can be greater than b2, how will a vertical graph be distinguished from a horizontal graph? The answer lies in this statement:

If the “x” coordinate is positive and the “y” coordinate is negative [y being subtracted from x], then the hyperbolas are HORIZONTAL. If the “y” coordinate happens to be positive and the “y” coordinate negative [x subtracted from y], then graphically, the hyperbola is VERTICAL.


This is where the English lesson came in. Mr. K insisted that it was “an” hyperbola since using an in a consonant was true in “an hour”. We realized that the English language is at times illogical and contradictory. If all of the rules were followed, we would either choose to have the silent letters or disregard them in the words: “talk”, “milk” and “walk”. Imagine what the Hulk would sound like. OMG Edward Norton as the new hulk. Unexpected, but I'm okay with it. By the way, you don't want to make Mr. K mad. He'll turn green [environmentally friendly?]

I'm such a subliminal message. But let’s not derail the train of thought here. The formation of a(n) hyperbola on a graph depends on a rectangular box, which forms a right angle triangle inside. The legs of the right triangle are composed of:

semi-conjugate + semi-transverse = c [the hypotenuse also equal to the length of the center (O) to a foci point.]

simply, this is where the Pythagorean [pie-tha-gore-eeiy-an] theorem is applied; a2 + b2 = c2


Then we spent the remainder of the class solving a series of hyperbolas in two different perspectives: graphically and symbolically.

To start we find the standard form by multiplying by (1/9) and (1/25) on each side
leaving us with [(x2 / 9) - (y2 / 25)] = 1

From this equation, we can derive the rest:

Center (0, 0) because nothing is being added to either of the x or y coordinates.

TRANSVERSE AXIS is found by finding the square root of 9 (a2) and multiplying by 2 (2a), in this case 2a = 6

CONJUGATE AXIS is found by finding the square root of 25 (b2) which is 5 (b) and then multiplying by 2. 2b = 10

VERTICES are the distances from the center to the vertex of both parabolas in the hyperbola. It helps if you know the length of the semi-transverse axis the coordinates of the vertices are (3, 0) and (-3, 0)

To find the FOCI, we start by recalling that they are the same distance away from the center as the hypotenuse of the right angle triangle formed by the semi-conjugate and semi-transverse axis serving as the legs of the triangle. Therefore we use a2 + b2 = c2.
In this case, 9 + 25 = c2
c2 = 34

Find the root of that and add/subtract from the x coordinate of the centre because this is a horizontal hyperbola, resulting in (root 34, 0) and (-root 34, 0)

Finally the asymptotes that help graph the two parabolas in a(n) hyperbola, it is important to recall grade 10 pre-cal where we spent time learning about oblique lines and how we find the slope. y = mx + b or simply "rise over run" and use a corner of the "box" made, meaning that we use the two legs of the triangle which happen to be the length of the semi-transverse axis and semi-conjugate axis. Rise 5, run 3 in both directions.
ASYMPTOTE = (5/3) and -(5/3)

Sorry if the post is messed up, but the library computers are timed... As for the next scribe.. I can't doodle, but nevertheless I dub with my very own schward the next scribe Rence.

PS. [post scribe, hahah] I know I said I was a hopeless apolitic when it comes to school elections and that I'd rather not vote. But I still encourage all of you to vote JOSEPH vice-prez.


Anonymous said...

On slide 7, shouldn't the equations of the asymptotes be:

y = ±5x/3

(Remember the y = mx + b form? In this case b = 0.)


y = ±5/3 <-- because this would result in a horizontal line.

&&Vote ZEPH for V.P. =)

m@rk said...

I need to agree with that anonymous. Hyperbolas don't have horizontal asymptotes. Good job for having a keen eye and keep up the good work.


PBnJamieSnagwich said...

thank you for noticing.... was really rushing and i'm lazy to change it LOL but it's always good to learn