Module 7: Conics Lesson 3 Notes Hyperbola Video 1 Transcript Hey everyone this is a video for module 7 lesson 3 hyperbola notes video 1.
So in the notes you found that the equation of a hyperbola there are 2 general forms. (!!!)!
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You have !! − !! =1, and then you have that with the sign on a different one. Notice that the sign in changing here, the “x” and the “a” just change position so the “x” term is negative in the bottom one and the “t” term is negative in the top one. When the negative sign is in front of the “x” term, the hyperbola will open up and down. When the negative sign is in front of the “y” term the hyperbola will open left and right. So you can see this one is in front of the “y” term, this is going to be left and right. And this one is in front of the “x” term, this one will be up and down. So up and down looks like this, left and right looks like this. !!
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Ok, so lets look at this example, its say for the hyperbola !" − ! = 1 find the following information. We need to find the coordinates of the center, this is found just like the circles, since I don’t have anything with my “x” and “y,” then my center is just going to be (0,0). You remember in the notes we are going to make a box, the coordinates of the corners of the box. So, I am going to go ahead and do that. So here is our equation, we have established our center as (0,0). The “x” squared term has an “a” squared term under it. If I take “a” squared and I just find “a.” I am going to come over here and go 5 to the right and 5 to the left. And my “b” squared term is 4 so my “b” term was 2. So that means I go up 2 and down 2. So what I do with these is, in the notes it told you to draw a vertical line through this one and a vertical line through this one, a horizontal here and a horizontal here. Where they connect that will form a box. I want to go ahead and just connect those with a rectangle. So there is the box that is going to guide me. Now to find those corners its just going to be at the points and its going to be (5,2), (-­‐5,2), (-­‐5,-­‐2) and (5,-­‐2)
Ok, so we found that one. The equations for the 2 asymptotes. So how you find those equations, is your going to find the line passes through here and that is going to be 1 of the lines for the asymptote. So I am going to find that equation first, I know that it passes through (5,2) and (0,0) . So I am going to quickly write the slope intercept form, given that the y-­‐intercept is 0. Then “y” equals my “m” is going to be the find the slope from here to here, I can go up 2 and right 5. So my slope is going to !
be !’s “x” plus 0, which I don’t have to write. So that is the equation of that line. So to find the equation of this other asymptote, as said in the notes the asymptotes go through the diagonals of the box that you formed. Ok, so I am trying to find the equation of these 2 lines. So I know the equation of this line, the y-­‐intercept is 0, I go !
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up 2 and left 5, which is going to be a − !’s slope. So my asymptotes are 𝑦 = ! 𝑥 and !
𝑦 = − ! 𝑥, asymptotes have the same definition here, the graph is going to get close to them but never touch them.
Now let talk about which way these are going to open, so I either have the option of opening here or here, or here and here. And once again that is determined by the sign, where the negative is, if the negative is in front of the “y” squared, I am going to go back to my form here, if its in front of the “y” squared its going to open left and right, this hyperbola is going to be here and here, its going to be to the left and right. Ok, now I have found the 2 asymptotes, does the 2 hyperbolas open up and down or left or right, it opens left, right. And the coordinates of the vertices, now, the coordinates of the vertices are these first dots that we put. So now that we determined that it opens left and right. Are vertices are going to be (5,0) and (-­‐5,0). So now I know roughly, very roughly, that here is a sketch of our graph. I know that these lines come close to these straight lines but never touch them because they are asymptotes. So I found the coordinates of the vertices. The direction of the transverse axis that’s just going to be horizontal it’s going to cut through here. So if its open left and right the transverse axis will cut horizontal. The coordinates of the foci are found by doing 𝑎! + 𝑏 ! = 𝑐 ! . So if I look at this formula, then my “a” squared equals 25 and my “b” squared equals 4, so to find “c” squared I would do 25 + 4. So that is going to be 29 equals “c” squared. To find the foci value I am going to take the square root of 29 and find the “c” value. So the square root of 29 is approximately 5.4, so “c” is approximately 5.4. And what you do with that value is use that as your foci, so your vertex is at (5,0), the foci is going to be (5.4,0) and on this side its going to be at (-­‐5.4, 0). The last step just ask you to sketch the hyperbola, that’s what we did throughout the problem and then you can graph the hyperbola in the calculator. I am going to take this and graph it in the calculator, you know you are going to have to solve for “y” to do that. So to solve that for “y” I have “x” squared over 25 minus “y” squared over 4 equals 1. I am going to multiply by the LCD, which would be 100. So if I multiply by 100 there. I would have 4x minus that would go to 25 and that would go to 100. To begin to get “y” by itself I would subtract “4x” squared from both sides. Then I would divide by negative so that’s going to change all my signs, divide by 25, and then take the square root. So “y” would equal plus or minus the square root of “4x” squared minus 100 over 25. I am not going to simplify that; I am just going to plug that in. That plus or minus is necessary to get the top and bottom half so I am going to have the square root of “4x” squared minus 100, so I have that numerator divided by 25. That closes the radical, so that’s my positive. Now I need the negative of that same value, and then I am going to hit graph. And that should give us about what our sketch gave us. So there is the top half, there’s the positive and there’s your negative. And that’s how you do that problem.