Precalculus (2018 edition)
Sal manipulates the equation 4y^2-50x=25x^2+16y+109 in order to find that it represents a hyperbola. Created by Sal Khan.
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- How do you know for sure if it's a conic? I mean, what if you try to simplify it and then you can't multiply it by a number so that it has the form ((x+a)^2)/b)+((y+c)^2)/d)=1? For example, what if, at3:58, the number on the right wasn't 100? What would you do then?(6 votes)
- It doesnt matter if it is some other number than 100. It can even be negative but not zero thought.(2 votes)
- Sal, In my upper level math class we are deriving the standard formulas for all the conic sections from the distance formula. I am stuck on the hyperbola. I can derive the equations for the asymptotes but I'm having trouble deriving the actual equation for the hyperbola. Can you help? Is there a video I can watch to give me some hint? thanks(3 votes)
Try this video on Proof of Hyberbola foci. https://www.khanacademy.org/math/trigonometry/conics_precalc/hyperbolas-precalc/v/proof--hyperbola-foci(5 votes)
- what are identity, contradiction and conditional equations?(3 votes)
AN EQUATION THAT HAS NO ANSWERS (NULL SET or NO SOLUTION
Subtract (3x) from both sides
AN EQUATION THAT IS TRUE BASED "ON THE CONDITION" THAT THE SOLUTION IS X=(A CERTAIN) THAT MAKES THE EQUATION TRUE
Subtract 3x from both sides AND Subtract 6 from both sides
The equation is true based on the condition that x=-11
AN EQUATION THAT HAS ALL REAL NUMBERS AS ITS ANSWERS. At some step in the solving of the equation you will get the same IDENTICAL terms on both sides of the equation.
3x+6=3x+6(Note the same terms on both sides)
COULD STOP HERE and say the solution is ALL REAL NUMBERS
If you Subtract 3x from both sides AND Subtract 6 from both sides you will get
A TRUE Statement
Solution ::::All Real Numbers
- Did Sal define what an asymptote was in an earlier video?(1 vote)
- It's the video:"Conic Sections: Intro to Hyperbolas : Introduction to the hyperbola" beginning about5:10
He doesn't explain them in details, but once the conic sections videos starts he mentions a couple of times, that those are important when dealing with hyperbolas. They are the lines to which the hyperbola gets closer and closer without ever touching it (see first and second video on conic sections)(5 votes)
- So, at1:24, Sal mentions that we can determine what kind of conic the equation will be by the x^2 and y^2 coefficients. WHY is it a hyperbolic equation if one coefficient is negative and the other is positive?(3 votes)
- it is just a characteristic of hyperbolic equations. This differentiates hyperbola from the ellipse's equation( having both x^2 and y^2 positive)(3 votes)
- Are the two asymptotes always intersecting?(2 votes)
- Yes, they always intersect at the center of the hyperbola, which is also the midpoint of the two focuses or the two vertexes.(4 votes)
- what do I do when there is no y^2?(1 vote)
- How do you tell if the equation is a hyperbola, parabola, ellipse, or circle? I understand how to solve them if I know which of the four it's asking for, but not when I need to differentiate it and solve it on my own.(2 votes)
- Ellipse is (x+h)^2 / a^2 + (y+v)^2 / b^2 = 1 where a and b determine the horizontal and vertical radii respectively, keeping in mind a is int he denominator under x and b is under y. h moves the ellipse left and right, where a negative h moves it to the right, and a positive h moves it left. Similarly a positive v moves it down and a negative v moves it up. Specifically h and v move the center of the ellipse, or more accurately where the radii are measured from. It's usually best to find this "center" point first. A shortcut is that the point is (-h, -v).
For this equation and all others, if it is not equal to 1 you can make it equal to 1, it just changes a and b. You need to have it equal to 1 though in this standard form. Say it is equal to c which is not 1 or 0, then you do this.
(x+h)^2 / a^2 + (y+v)^2 / b^2 = c divide both sides by c
(x+h)^2 / ca^2 + (y+v)^2 / cb^2 = 1
So now the two radii are not a and b, but sqrt(c)a and sqrt(c)b
A circle is a special version of an ellipse, basically where a = b. let's call this c. (x+h)^2 / c^2 + (y+v)^2 / c^2 = 1. In this form you can change it to the more normal form of a circle.
(x+h)^2 / c^2 + (y+v)^2 / c^2 = 1
1/c^2 [(x+h)^2 + (y+v)^2] = 1
(x+h)^2 + (y+v)^2 = c^2
So c is the radius all around.
A hyperbola is also similar in equation to an ellipse, there is just a minus sign in the equation. Just one minus though. So that means a hyperbola is either (x+h)^2 / a^2 - (y+v)^2 / b^2 = 1 or (y+v)^2 / b^2 - (x+h)^2 / a^2 = 1. If it's -(y+v)^2 / b^2 - (x+h)^2 / a^2 = 1 with two minuses on the left or (y+v)^2 / b^2 + (x+h)^2 / a^2 = -1 where there is no minus on the left but equal to -1 on the right there is no solution.
A parabola meanwhile is an equation where only one variable is squared. so you could put it into the form x = ay^2 + by or y = cx^2 + dx. keep in mind though b and d can be 0, and there are no powers above 2.
Does this answer your question?(2 votes)
- At10:03shouldn't it be y=3 and y=-7?(2 votes)
- Around8:42, Sal talks about whether the graph is horizontal or vertical. Can a shortcut be just that when the y^2 value is positive, the graph is vertical and when the x^2 value is positive, the graph is vertical? Or is that not always accurate?(1 vote)
- Other than your editing error (when the x^2 value is positive, the graph is horizontal), that is generally true. It comes down to the question "Am I trying to take the square root of a negative number?"
The exceptions I can think of are the sorts of things you won't run into at this level. Trigonometry at the earliest, and maybe not even then.(2 votes)
Let's do another conic section identification problem. So, I have 4y squared minus 50x is equal to 25x squared plus 16y plus 109. So, the first thing I like to do is to group all of the x and y terms onto one side of the equation and leave all the constants on the other side. So let's do that. So, on the left-hand side I'll put the 4y squared. 4yy squared. And, actually, I'm also going to group all the x and y terms in this step. So the 4y squared. Let's move this 16y onto the left-hand side. So if I subtract 16y from both sides of this equation, I get minus 16y, minus 16y on the left-hand side and of course it will disappear on the right-hand side. And then I want to subtract the 25x squared from both sides of this equation. So I get minus 25x squared minus 50x. That's that right there. And then I'll leave this 109 on the right-hand side. It's equal to 109. And now that we have the x's and the y's on the same side of the equation, we know what type of -- we know the general direction we're going to go in. Because they're on the same side. They have different coefficients. And one is positive and one is negative. So that lets us know that we're dealing with a hyperbola. So let's complete the square and get it into the standard form. So, the easiest way to complete the square is if you have a 1 coefficient on the y squared and the x squared terms. So let's factor out a 4, in this case. So you get 4 times y squared minus 4y. I'm going to add something later, when I complete the square. Minus 25 times x squared, plus, let's see, minus 50 divided by minus 25 is 2. Plus 2x, I'm going to add something later. Is equal to 109. And the things we're going to add, those are what complete the square. Make these things a perfect square. So, if I take this, you have a minus 4 here. I take half of that number. This is just completing the square, I encourage you to watch the video on completing the square where I explain why this works. But I think I have a minus 4. I take half of that, it's minus 2. And then minus 2 squared is plus 4. Now, I can't do one thing to one side of the equation without doing it to the other. And I didn't add a 4 to the left-hand side of the equation. I actually added a 4 times 4, right? Because you have this 4 multiplying it out front. So I added at 16 to the left side of the equation, so I have to also add it to the right-hand side of the equation. Right? This is equivalent to also having a plus 16 here. That might make a little bit clearer, right? When you factor it out, and it becomes a 4. And we would have added a 16 up here as well. Likewise, if we take half of this number here. Half of 2 is 1. 1 squared is 1. We didn't add a 1 to the left-hand side of the equation, we added a 1 times minus 25. So we want to put a minus 25 here. And, likewise, this would have been the same thing as adding a minus 25 up here. And you do a minus 25 over here. And now, what does this become? The y terms become 4 times y minus 2 squared. y minus 2 squared. Might want to review factoring a polynomial, if you found that a little confusing, that step. Minus 25 times x plus 1 squared. That's that, right there. x plus 1 squared, is equal to, let's see, 109 plus 16 is 25 minus 25, it equals 100. We're almost there. So we want a 1 here, so let's divide both sides of this equation by 100. So, you will get y minus 2 squared. 4 divided by 100 is the same thing as 1/25, so this becomes over 25. Minus, let's see, 25/100 is the same thing as 1/4, so this becomes x plus 1 squared over 4 is equal to 1. And there you have it. We have it in standard form and, yes, indeed, we do have a hyperbola. Now, let's graph this hyperbola. So the first thing we know is where the center of this hyperbola is. Is the center of this hyperbola is at the point x is equal to minus one. So it's an x is equal to minus 1. y is equal to 2. And let's figure out the asymptotes of this hyperbola. So if this was -- this is the way I always do it, because I always forget the actual formula. If this was centered at 0 and it looked something like this. y squared over 25 minus x squared over 4 is equal to 1. I do this to figure out what the asymptotes would have been if we were centered at 0. Because it's a lot easier to deal with these equations than to deal with these. So we could solve -- we multiply both sides by 100. We're kind of unwinding what we just did. So if you -- actually, let's multiply both sides by 25. So then you get y squared minus 2 over 4x squared is equal to 25. And then, I'll just go right here. And then if I add 25 over 4x squared to both sides, I get y squared is equal to 25 over 4x squared plus 25. And so, y is equal to the plus or minus square root of 25 over 4x squared plus 25. And like always, the asymptotes, the hyperbola will never equal the asymptotes or intersect the asymptotes, but it's what the graph approaches as x approaches positive and negative infinity. So, as x approaches positive and negative infinity, and you'll learn the concept of limits later on. But I think you get it at this point. Because that's what the idea of an asymptote even is. Is that as x gets really larger, approaching this line -- so as x approaches positive or negative infinity, as we've done in the previous videos, this term starts to matter a lot less. Because this term is huge. So then y is approximately equal to plus or minus the square root of just this term. Now, the square root of just this term is 5/2 x. So those would be our asymptotes if we were centered at 0. But, of course, we're centered at negative 1, 2. So let's graph that. And then we could figure out if it's a upward-opening or downward-opening graph. We're centered at negative 1, 2. So I want to be in the, that's my y axis. This is my x axis. And we're centered at minus 1, 1, 2. That's the center. And this would've been the two lines of the asymptotes if we were centered at 0. But now this tells us the slope of the two asymptotes. So the asymptotes are going to intersect at the center of our hyperbola, so to speak. So these are the slopes of the two asymptotes. And one is positive 5/2. So, positive 5/2 means if we go over 2, so 1, 2, in x, we go up 5. So, 1, 2, 3, 4, 5. So we'll end up right over there. So I can draw that line, I just need two points for a line. So that line would look like that. And the other asymptote is minus 5/2. So for every 2 we go over to the right, we go down 5. So, 1, 2. 1, 2, 3 4, 5. So we end up right about there. And so that line would look like that. Good enough. So those are the two asymptotes, and they go on forever in those directions. And now we can think of it two ways. We could either say, OK, if we look at -- actually, look at this one. If it was centered at 0, could x equals 0? Well, sure x could equal 0. If x is 0, then y squared over 25 equals 1. y squared equals 25. y would be plus or minus 5. So, in this case, this term could be equal to 0. So we could say that x could equal negative 1. If x is equal to negative 1, then y minus 2 squared over 25 will equal -- let's do that. Let's set. If x is equal to negative 1, x is equal to negative 1, then what does this expression become? I don't want to lose it, so I'll write it right there. So then you get y minus 2 squared over 25. This becomes 0 minus 0 is equal to 1. So you get y minus 2 squared over 25 is equal to 1. y minus 2 squared is equal to 25. Just multiplied both sides by 25. y minus 2 is equal to plus or minus, I'm just taking the square root of both sides, 5. So y minus 2 is equal to positive 5 or y minus 2 is equal to minus 5. Add 2 to both sides of this, you get y is equal to 7. Add 2 to both sides of this, you get y is equal to minus 3. So we know that the points minus 1, 7 and minus 1, minus 3, are both on this graph. So, minus 1 is here. 1, 2, 3, 4, 5, 6, 7, minus 1, 7, and minus 1, 1, 2, 3, are both on this graph. So that lets us know, since we're inside here, this tells us that this is kind of a vertical asymptote, and another way to guess it is, if you see that y squared term is positive. Or the other way to think about it is, when you take the positive square root, when you take the positive square root, you're always going to be a little bit above the asymptote. That's the other way to think about this. That we're always going to be a little bit -- and this is the positive square root. The positive square root is the top line. So we're always going to be a little bit above the asymptote. This is the asymptote. But we're always a little bit above it. And obviously as this number gets larger, this starts to matter a lot less, so the graph is going to look something like this. It's going to come down and then go off, and never quite touch the asymptote, but approach it. So it's going to get really close to the asymptote, and then go off, and go off in that direction. Anyway, hope you found that helpful. This was a slightly hairier problem, so it should be instructive.