The result of our folding is a hyperbola. After we found this out we went on a quest to find any patterns within, just like what we did with the ellipse.
Mr.K told us to pick a point anywhere on the hyperbola and find the distance from F1 to P and F2 to P. We then take the difference of these two numbers. We did this several times and found out that the difference was constant no matter where you put the point. We also found out that this constant value is exactly the same as the distance between the two vertices's of the hyperbola.
The distance between the two vertices's is known as the Transverse axis. The conjugate axis is the perpendicular bisector of the transverse axis. This intersection point between the transverse axis and the conjugate axis is known as the center.
These are notes you will need to understand so you can draw these graphs with ease. The technical terms are not all that important.Below is the standard form for the Hyperbola equations along with the similarities and differences between them.
We are now back to the anatomy of the hyperbola.
Just like the ellipse, this also has a Pythagorean triangle within it. The distance of the hypotenuse is the same distance as the distance between the center and one of the focus points. With this new knowledge we can create hyperbolas with limited amounts of information.HOMEWORK!
(i) We did the standard form equation in class as it shows on this slide. If you need help understanding, what we did was take that 225 on the right side of the equation and divide by that number throughout the whole equation. This will help us get the preferred number 1. After we just simplify the equation and get...(x^2 / 9) - (y^2 / 25) = 1
(ii) The transverse axis equals 2a. We can find a from the number under x^2. If you noticed this term is positive therefore making this equation a horizontal one. a^2 is under the x^2 in this type of graph. So if we take the square root of 9 we can find a. So a=3. Therefore the transverse axis is 6.
Now to find the conjugate axis's length. The conjugate axis is 2b. Under the y^2 term we can find b^2. If we take the square root of 25 we will get what b equals, multiply by 2 and the conjugate axis's length is 10.
In order to find the coordinates of the foci and the vertices's we must know where the center is. Since the x^2 term and the y^2 term have nothing within it to shift the graph; the center is (0,0). So the vertices's is the distance of away from the center. Therefore the vertices's must be (3,0) and (-3,0). To find the foci we must find c. Which requires the Pythagorean theorem.
a^2+b^2 = c^2 will be used in this case. (Remember it is not always a^2 + b^2 = c^2, it is actually (leg1)^2 + (leg2)^2 = (hypotenuse)^2) With this step we get c to be a value of root(34). Therefore the foci are located at (root(34),0) and (-root(34),0).
Finding the asymptotes. I found the asymptotes by drawing the rectangle as done when constructing a hyperbola. When I found the corner points I was able to figure it out. The points I got were (3,5) , (3,-5) , (-3,-5) , (-3,5). In addition all of the asymptotes always intersect with the origin. So I used (3,5) and (0,0) to find the positive asymptote and (-3,5) and (0,0) to find the negative asymptote. So with these points I just plugged it into the two point formula which is...
(y2-y1)/(x2-x1) = (y1-y0)/(x1-x0)
This is a formula we learned way back in grade 10 but it's still useful. I ended up getting two asymptotes that were y=5/3x and y = -5/3x. There are probably different ways but I chose this way.
The graph should look something like this.
a^2+b^2 = c^2 will be used in this case. (Remember it is not always a^2 + b^2 = c^2, it is actually (leg1)^2 + (leg2)^2 = (hypotenuse)^2) With this step we get c to be a value of root(34). Therefore the foci are located at (root(34),0) and (-root(34),0).
Finding the asymptotes. I found the asymptotes by drawing the rectangle as done when constructing a hyperbola. When I found the corner points I was able to figure it out. The points I got were (3,5) , (3,-5) , (-3,-5) , (-3,5). In addition all of the asymptotes always intersect with the origin. So I used (3,5) and (0,0) to find the positive asymptote and (-3,5) and (0,0) to find the negative asymptote. So with these points I just plugged it into the two point formula which is...
(y2-y1)/(x2-x1) = (y1-y0)/(x1-x0)
This is a formula we learned way back in grade 10 but it's still useful. I ended up getting two asymptotes that were y=5/3x and y = -5/3x. There are probably different ways but I chose this way.
The graph should look something like this.
Sorry for my poor drawing xD. Well that's about it. If any information on here is incorrect I will try to change it as soon as possible.
The next scribe will be Aldrin S!!! WOOT! show us your mad skills xD.
