2. Algebraically, a hyperbola's equation can either look like this:
. Or it can look something like this: 
. (I will go into further detail of what all of these variables mean later on.) If you were to graph these equations, a hyperbola would look like two parabolas opposite of one another, facing in different directions. Because there are two ways the equation can be written, the hyperbola may appear differently as well, depending on which of the two terms ("x" or "y") is negative. If your "y" term is subtracted, your hyperbola will branch to the left and right. If your "x" term is subtracted/negative, your hyperbola will branch up and down.  |
| "y" term is negative (http://tutorial.math.lamar.edu/Classes/Alg/Hyperbolas.aspx) |
 |
| "x" term is negative (http://tutorial.math.lamar.edu/Classes/Alg/Hyperbolas.aspx |
When discussing hyperbolas, we need to remember these features: center, transverse axis, conjugate axis, vertices, co-vertices, foci, eccentricity, and asymptotes. To begin, to find the center, we need to first remember something very crucial: "h" goes with "x" and "k" goes with "y". Knowing this, we can look at our standard form and find our "h" and our "k", basically allowing us to find our center (h,k). You can plot the center, and go on from there.
Moving on, looking at your standard form, whichever term ("x" or "y") comes first will determine whether your transverse axis is horizontal or vertical. If "y" is the first term, that tells us that our transverse axis will be vertical. If "x" is the first term, that tells us that our transverse axis will be horizontal. Knowing this, we will know that our conjugate axis must be the opposite. So if our transverse axis is vertical, our conjugate axis will be horizontal and vice versa. (Side note: By knowing if the axes are horizontal or vertical, we will know if our lines will be "x=" or "y=". And because we know our axes go through our center, we know those "x=h" and "y=k". Less work for us!)The distance of the transverse axis is 2a ("a" is the square root of the denominator beneath the first term) and the distance of the conjugate is 2b ("b" is the square root of the denominator beneath the second term). If you want to know which lines are your transverse and conjugate axes just by looking at a graph, take a look at the branches. If your branches go left and right, you know your transverse axis is horizontal. If your branches go up and down, you know your transverse axis is vertical. If you want to know the length of either of the axes, simply count the units (to find "a" or "b", divide the entire length of the axis by 2).
Going on to vertices, these are fairly easy to find. Graphically, the vertices are basically the ends of the transverse axis. Look at the end points and you have the vertices. Algebraically, all you need to know is what your "a" is. If your transverse axis is "x=" then you know that the "x's" in your vertices will remain the same as your center. From there, all you need to do is add and subtract "a" to "k". {ex: center: (0,0); trans: x=0; vertices: (0,0+a) & (0,0-a)} The same can be said for the co-vertices. Look for the end points of the conjugate axis on your graph. Algebraically, find "b" and add that to the value that is changing ("x" or "y")(HINT: If your "y" values changed for your vertices, your "x" values will change for your co-vertices). Your foci is the same idea, with the adding to either the "x" or "y" values of your center (because your foci are "c" units away from your center, correct?). You find "c" by using this formula: a2 + b2 = c2. After finding "c", you add that to the SAME VALUE ("x" or "y") that you added for your vertices because your foci lay on the same line (transverse axis) as your vertices. So if your "x" values did not change for your vertices, then your "x" values will not change for your foci. You will take the "k" value of the center and add and subtract "c" from it to find both foci. To plot, your foci will be going the same direction as your vertices, and further than your vertices in fact (I will explain soon). If your transverse axis is horizontal, your foci will be going horizontal.
After all of this, now you need to calculate the eccentricity, how much the conic deviates from the shape of a circle. To find eccentricity, you divide "c" by "a" (c/a). The foci has a great effect on the eccentricity. The greater the foci is (as in the farther it is from the center), the more it deviates from being a circle, the greater the eccentricity is. The greater your eccentricity is, the wider it is; the less your eccentricity is, the "skinnier" it is.
Finally, let's talk about asymptotes (they are the diagonals of the box we create from the conjugate and transverse). We have given formulas to use to find them:
(used when transverse is horizontal) and 
(when transverse is vertical). You take your variables and plug them in. However, you can also use our knowledge from algebra 1 and use our slope-intercept formula: y=mx+b. Our slope (m) would be either a/b (if our transverse is vertical because our "a" is our "y"/ our "rise", and our "b" is our "x"/ our "run") or b/a (if our transverse is horizontal). From there, you plug in our center (x,y) (the center goes through our lines) and find 'b". REMEMBER, because we have two asymptotes, one slope should be POSITIVE a/b or b/a (it depends on the graph) while the other should be the NEGATIVE of that. You can also count and find the slope on the graph.  |
| hyperbola and its features (https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjToBqnI6fP8DDSxy9Ae6udUE1V6aukjRmQAEfbJTcyZaIN1HRvNd8ukmy6sFXLG0RpCuH7hxXR_JEpfTzaU5dr1on3yfIg9bbifyNswsOSSr2Yk3HnLanhiGEEwpKooLBBA1k80elZBQg/s1600/kirch.png) |
3. Real world applications of hyperbolas can be seen in cooling towers for nuclear reactors. These towers were built into the shape of hyperboloids (hyperbolas in 3D!) on purpose. When building these, there were problems in creating a building that would be able to withstand high winds and intense conditions without using that much material. Finally, engineers discovered the benefits of creating these towers into the shape of a hyperbola.
By building these towers into hyperboloids, they are able to stand up to high winds and use less material. "A 500 foot tower can be made of a reinforced concrete shell only six to eight inches wide." (http://www.pleacher.com/mp/mlessons/calculus/apphyper.html) Also, they help the upward air flow move faster, cooling things faster and better. Before, these towers were built in a cylindrical form but they had encountered problems with the shape.
Works Cited:
1. http://www.purplemath.com/modules/hyperbola.htm
2. https://people.richland.edu/james/lecture/m116/conics/hypdef.html
3. http://www.structuremag.org/article.aspx?articleID=326
4. http://www.pleacher.com/mp/mlessons/calculus/apphyper.html
5. http://www.youtube.com/watch?v=Z6cwpsDC_5A
6. http://tutorial.math.lamar.edu/Classes/Alg/Hyperbolas.aspx
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