Pre Calculus 8.3 - Hyperbolas

Summary

In this Pre-Calculus session, John Swanson explores hyperbolas, one of the four conic sections. This lecture is all about analyzing hyperbolas' equations, graphing them, solving application problems, and systems of non-linear equations. Swanson draws parallels between hyperbolas and ellipses, detailing their similar properties but emphasizing critical differences, like the concept of asymptotes, which are unique to hyperbolas. Detailed instructions for constructing hyperbolas, complete with vertices, co-vertices, and asymptotes, are provided. The lesson also includes solving systems of equations involving hyperbolas through graphing, substitution, and elimination methods, offering a comprehensive understanding of these fascinating mathematical structures.

Highlights

• Hyperbolas are the fourth type of conic sections after ellipses, parabolas, and circles 🔄.
• The equation for hyperbolas swaps addition for subtraction in contrast to ellipses ➖.
• Hyperbolas have two types: horizontal (left-right) and vertical (up-down) orientations ⬅️➡️️.
• The transverse axis in hyperbolas can be shorter or longer, varying from ellipses ↕️.
• Hyperbolas need a rectangle formed by vertices and co-vertices to sketch asymptotes sharply 📐.
• Graphical solutions can visibly demonstrate hyperbolas interacting with other geometric figures effectively 🌈.
• Comparing and graphing hyperbolas together with ellipses aids in recognizing their symmetries and asymmetries 🚦.
• Solving non-linear equations by different methods provides clarity on how hyperbolas behave in models ⚙️.
• Finding constant differences helps define hyperbolas precisely during equation manipulation 🧮.
• John Swanson offers detailed guidance on solving practical problems involving hyperbolas efficiently 🔍.

Key Takeaways

• Hyperbolas differ from ellipses and parabolas due to their asymptotic nature 📏.
• Graphing hyperbolas involves identifying vertices, co-vertices, and drawing asymptotes 📉.
• Key differences: in hyperbolas, the foci are outside the vertices, contrary to ellipses📍.
• To solve systems of equations with hyperbolas, use graphing, substitution, or elimination methods. 🌐
• Equation manipulation uses constants and properties like 'a² + b² = c²' for hyperbolas.
• John Swanson highlights the mathematic similarities and differences of conic sections to simplify complex graphs 🎨.

Overview

Hyperbolas, characterized by their asymptotic nature, provide unique insights into the world of conic sections. Unlike parabolas, which continuously curve, hyperbolas approach asymptotes, offering a bridge between the world of ellipses and parabolas. In this lecture, John Swanson leads us through the wondrous intricacies of hyperbolas, discussing their properties, graphing strategies, and applications in real-world problems.

Swanson methodologically explains how hyperbolas relate to ellipses by comparing their defining principles and mathematical equations. Both conic sections are visually and algebraically linked, yet through subtraction rather than addition in their fundamental formulas. Details like distinguishing the transverse and conjugate axes form a critical part of understanding hyperbolas' layout in a graph.

A significant part of the lecture involves practical problem-solving where Swanson walks through the substitutions and eliminations needed to solve equations involving hyperbolas. Understanding these mathematical tools is vital for students to transfer abstract mathematical concepts into tangible applications seen in fields such as physics, engineering, and computer science.

Chapters

• 00:00 - 00:30: Introduction to Hyperbolas In the 'Introduction to Hyperbolas' chapter, the discussion focuses on hyperbolas as a type of conic section. Though they resemble parabolas, hyperbolas share more characteristics with ellipses. The learning objectives include analyzing the equations representing hyperbolas, graphing them, and finding equations for hyperbolas based on specific conditions.
• 00:30 - 01:30: Comparison with Ellipses In this chapter, the focus is on comparing ellipses with hyperbolas and solving related mathematical problems. The chapter begins by addressing application problems that involve hyperbolas and systems of non-linear equations, building upon concepts introduced in the previous chapter. A comparison is drawn between ellipses and hyperbolas, emphasizing their similarities and differences. The definition of ellipses is discussed as a set of points where the sum of distances from two fixed points remains constant, setting the stage for further exploration of hyperbolas and other mathematical concepts.
• 01:30 - 02:30: Equation of a Hyperbola The chapter titled 'Equation of a Hyperbola' presents the mathematical concept that defines a hyperbola. It starts off by explaining that a hyperbola is a set of all points where the difference of the distances from any given point to two fixed points (foci) is constant. The text elaborates on the core idea, emphasizing the difference, instead of a sum as seen in ellipses, which is constant. It touches upon the formula involving 'm' and 'n' where the expression 'm minus n' equals a constant value. This underlines the importance of absolute values in maintaining positive differences between distances.
• 02:30 - 05:30: Hyperbola Structure and Graphing The chapter discusses the structure and graphing of hyperbolas. It explains that the general equation for a hyperbola is similar to that of an ellipse, where instead of adding the terms, they are subtracted. Specifically, the equation for an ellipse is x²/a² + y²/b² = 1, while for a hyperbola, it is x²/a² - y²/b² = constant. This subtraction results in two separate curves or 'arms' of the hyperbola.
• 05:30 - 10:30: Finding Equations for Hyperbolas This chapter covers the topic of hyperbolas, focusing on how to find equations for them. It explains that unlike ellipses, where the focal length (c) is determined by the equation a^2 - b^2 = c^2, hyperbolas require adding a^2 and b^2 to equate to c^2, aligning with the Pythagorean theorem. Additionally, it notes the distinctive feature of hyperbolas where the foci are located outside of the vertices.
• 10:30 - 29:00: Solving Systems of Equations The chapter discusses solving systems of equations, focusing on the properties and characteristics of hyperbolas. It distinguishes between horizontal and vertical hyperbolas, explaining their orientations. Horizontal hyperbolas open left and right, while vertical ones open up and down. Importantly, hyperbolas are not considered functions due to their multiple outputs for single inputs.
• 29:00 - 30:00: Conclusion and Encouragement The concluding chapter reaffirms the understanding of geometric structures, focusing on the horizontal and vertical openings, foci, and vertices. The discussion highlights that while the foci are outside the vertices, similar to a parabola, yet remain within the hyperbola's curve. The vertices are found exactly at the expected turning points of the hyperbola, reinforcing the lesson's key learning objectives on the structural dynamics of hyperbolas. This serves as an encouragement to appreciate the beauty and complexity of mathematical shapes.

Pre Calculus 8.3 - Hyperbolas Transcription

• 00:00 - 00:30 welcome back to section three of chapter eight this section we're going to look at the fourth type of conic section which is the hyperbolas they look like parabolas kind of but they actually have a lot more in common with ellipses believe it or not are learning targets today we're going to analyze the equation of hyperbola graph hyperbolas find an equation of hyperbola satisfying given conditions
• 00:30 - 01:00 solve application problems involving hyperbolas and solve systems of non-linear equations which we actually did a little bit of last chapter so um first of all looking at ellipses and hyperbolas i said they were very similar so let's take a look to compare them ellipses the definition is a set of all points where the sum of the distances from the point to two fixed points is constant hyperbola definition
• 01:00 - 01:30 is the set of all points where the difference of the distances from the two from the point to two fixed points is constant so the difference here is whether it's a sum or the difference is it m plus n or is it m minus n is a constant so we have a point that distance minus that distance and then absolute value because the positives because
• 01:30 - 02:00 gives us one arc the negatives gives us the other um in their arms so that equals just a constant whatever the constant is for that particular hyperbola ellipses the equation was x squared over a squared plus y squared over b squared equals one that's the general equation for a hyperbola basically the same thing except instead of adding we're subtracting x squared over a squared minus y squared
• 02:00 - 02:30 over b squared equals one um ellipses to find that focal length that's c it was a squared minus b squared equals c squared it was very similar to the pythagorean theorem except we're subtracting for hyperbolas it actually is the pythagorean theorem we're adding those two things um notice in hyperbolas the foci are on the outside of our vertices whereas the
• 02:30 - 03:00 ellipses they were on the inside so let's take a closer look at some of these hyperbolas we have horizontal ones and we have vertical ones just like all the other conic sections the horizontal ones are the ones that open left and right they look kind of like that the vertical ones are the ones that open up and down neither of these two things are functions i was about to call one of them a function but that doesn't work out
• 03:00 - 03:30 but the horizontal again open left right vertical open up and down we have foci the foci again are outside the vertices but they're inside kind of like the parabola they're inside the curve of the hyperbola we have vertices the vertices are right where you'd expect them to be right at the turning points of the hyperbola i'm
• 03:30 - 04:00 kind of like with a parabola except we have two of them we have co-vertices now these co-vertices kind of like with an ellipse has co-vertices but notice these co-vertices aren't actually even on the hyperbola they're not on the graph at all they do have meaning we're going to use them and see what they are um but they're not actually part of the graph which is kind of weird
• 04:00 - 04:30 we have a transverse axis that in an ellipse would be the major axis it's the one that goes in between the two vertices in an ellipse the major axis was the longer of the two that is not necessarily the case in a hyperbola it could be much shorter but it's the one that connects the two vertices the conjugate axis is going to be the
• 04:30 - 05:00 one that connects the two co-vertices it'll be along that line of symmetry well i guess they're both along a line of symmetry but it would be the one that separates the two arms so the conjugate axis goes between the two covertices and then if we connect those vertices and co-vertices with a rectangle we can draw lines through
• 05:00 - 05:30 the corners those are asymptotes one of the big differences between hyperbolas and parabolas is hyperbolas are asymptotic parabolas are not parabolas continually curve up hyperbolas go towards those asymptotes and again to find the asymptotes we just we make a box using the vertices and co-vertices that's where we use those co-vertices
• 05:30 - 06:00 and then connect the corners so we do it with the vertical ones as well and that will help us to draw these hyperbolas we have distances that horizontal distance a just like with an ellipse the vertical distance is b just like with an ellipse
• 06:00 - 06:30 the uh the book kind of switches it just like with an ellipse i like keeping the horizontal distance as a the vertical distance is b um the focal distance is going to be c just like with an ellipse and we already saw that a squared plus b squared equals c squared in this case um our equation if it's centered at 0 0 for horizontal it's x squared
• 06:30 - 07:00 over a squared minus y squared over b squared equals one um if we wanted to move the vertex we have x minus h y minus k just like we have for every single other conic section um and line and everything else taylor can over your bus has returned to pick you up taylor can over your bus has returned
• 07:00 - 07:30 that's horribly irritating um so we have our h or k um and that moves the vertex or the uh the center um for vertical we get the same things the difference between horizontal and vertical it's not which one's bigger the a squared of the b squared it's which one's positive uh horizontal we have x squared minus y squared vertical we have y squared minus x
• 07:30 - 08:00 squared that is the difference so which one's positive tells you which way it's going um we have a squared plus b squared equals c squared um the asymptotes are if it's centered at 0 0 the line's y equals plus or minus b over a times x because to go from corner to corner it goes to the center which is zero zero
• 08:00 - 08:30 we go up b we go over a the slope is always b over a and that's true here as well as long as you don't switch the b and the a's if we wanted to move the center off of the origin we just change the x and y to x minus h and y minus k point slope form that's all it is and it's plus or minus it's both that gets us both of our lines um
• 08:30 - 09:00 yep those are the asymptotes already got that all right so what are we going to do with these hyperbolas find the constant difference just like we have with the ellipses and things like that so we have our two foci negative 13 0 and 13 0 and we have a point 5 comma 0. so difference equation negative 13 minus 5 squared plus 0 minus 0 squared minus because it's a difference root 13 minus 5 squared
• 09:00 - 09:30 plus or plus 0 minus 0 squared we can um combine our like or do the subtraction so negative 13 minus 5 is negative 18 squared and then plus 8 squared when we square and square root those we're going to get 18 minus 8 which is 10. so the constant difference is going to be 10. write equations in standard form we have
• 09:30 - 10:00 center vertex and focus so the vertex is going to give us either our a or our b value from the center the center gives us our h in our k the focus gives us a c value allowing us to find the other one being that this one is always changing the y and it's a vertex that means it's going to be a vertical we're going to have y squared over b squared minus x squared over a squared
• 10:00 - 10:30 equals 1. plug in what we know the b is 5 so we can plug that in to get 5 squared and then we can use the 5 and the 13 to figure out what that a is going to be so we can square those we can subtract a squared is 144. so people like oh a is 12 all right if we're graphing it that matters if we're not graphing it we need a squared in the equation anyway 144. um so we have y squared over 25 minus x
• 10:30 - 11:00 squared over 144 equals one notice this one is um that transverse axis is much smaller than the conjugate axis that's okay we'll see what that looks like when we start graphing um here's a graph write an equation from the graph so we have a center is zero zero we have our distance out to the vertex is going
• 11:00 - 11:30 to be eight we have a distance to the co-vertex is going to be five this is horizontal so it'll be x squared minus y squared so we have x squared over a squared minus y squared over b squared a is eight no seven a is seven um y is five or b is five so we can plug those in and square them x squared over 49 minus y squared over 25 equals 1.
• 11:30 - 12:00 i'm graphing hyperbolas you know when i say they're very similar to ellipses i mean they're very similar to ellipses so if we have x squared over 36 minus y squared over 49 equals one we find the vertices negative six zero and six zero because the center is at zero zero we find the co vertices zero and seven the asymptotes if we need an equation for an asymptote which it does say we need to find the
• 12:00 - 12:30 asymptotes um we have that plus or minus b over a times x and then graphing it so to graph this we start with our center and then we're going to go our vertices in this case are horizontal 2 3 4 5 6. make a point come on pen
• 12:30 - 13:00 there we go there's a point two three four five six our co-vertices up and down seven two three four five six seven we're going to
• 13:00 - 13:30 make the box connect for asymptotes and then we're going to sketch the hyperbola in this case it's horizontal so
• 13:30 - 14:00 it's going to be here and here if it were vertical we would have done exactly the same thing except we would just sketch it on the top and the bottom so let's see how close i was and i hope that i used one to one on this one i guess we're about to find out all right so i'm plotting my points there's the box okay i did use one to
• 14:00 - 14:30 one you can barely see it because my drawing is so close some asymptotes and then sketching it wow the the right one was really good my left one i missed a bit but my right one was really good so that's the process there so let's do it again but this time we have a different center still horizontal but our center is now 4 3 our vertices from 4 3 we're just going
• 14:30 - 15:00 to go over 3 each way so we get 1 3 and 7 3. unless you have to find what those values are you don't need to find them because you can just start at your center and go over 3 over each direction and same thing up or down eight our asymptotes that would be equations of the asymptotes it's the plus or minus b over a and then we have x minus h and y minus k notice that the y minus three is there the x minus four is there like those pieces go straight into that
• 15:00 - 15:30 equation so we have center four three one two three four one two three our center and then vertices let me go over three over three one two three four five six seven go up and down eight make our box
• 15:30 - 16:00 make asymptotes and then graph it and again it's horizontal so we're going there if it was vertical you'd done everything the same except that it would have been here and here
• 16:00 - 16:30 that would be the difference with vertical but it's not vertical so we're not doing the purple um let's figure out how close i was so we have the center we have some vertices which i was right on in those we have our box we have asymptotes asymptotes and drawing sketching it again i'm close on the right ones but the left ones i'm struggling with today
• 16:30 - 17:00 but again i am not an artist i don't expect you to be an artist either so that's hyperbola so what about solving these systems of equations that are not linear so for solutions we have different solutions or different types of equations when we have lines we can have two lines they intersect once they could be parallel
• 17:00 - 17:30 and not intersect at all so we could get zero solutions they could also be the same line so we could have an infinite number of solutions so with lines we have the option of 1 zero or infinite solutions when we have things other than lines we get a lot more options though if we have a line and a parabola we could have two solutions see there's two intersection points
• 17:30 - 18:00 there we could have one where the line is tangent to the parabola or we could have zero where it misses the parabola entirely so there could be two one or zero and that's what the line and a parabola with an ellipse and a parabola or an ellipse a circle in a parabola we could have 2 we could have 1 where it's just tangent and that could be a few different places
• 18:00 - 18:30 we could have none if it's um just outside of it and that could happen all sorts of places in fact there for a second we had none where it went around now this is why i work from home now we have two and now we have four we could get a total possible four so we could have zero one two three or four solutions so here's three
• 18:30 - 19:00 i forgot that we hadn't even done that one yet so zero one two three or four possible solutions and then if we have like circles and ellipses or hyperbolas and ellipses it's the same kind of thing we can get up to
• 19:00 - 19:30 four solutions anywhere from zero to four um so solving by graphing we would graph just like we would always solve by graphing we graph it and we see what it is first we have to get it into y equals form unless you're using desmos as most you can plug in it in however you want then we get our graphing calculator and we're gonna enter them in so y equals we have negative three x
• 19:30 - 20:00 and we have negative three x squared plus six and we graph it we have our see we have our line thinking and the parabola and here we can see that we have two intersection points
• 20:00 - 20:30 and so to find them we can go to calc which is second trace and we would find the intersection point five when when we do this it's going to ask us some questions the first is the first curve so it'll be on y1 instantly so we can just hit enter then it'll ask us for the second curve it automatically jumps to y2 hit enter and then it guess this is where we choose which one we're looking for so just get close to one of them and hit enter and it'll give you the
• 20:30 - 21:00 intersection at negative one comma three to do the other one you do the exact same thing except when it's going to ask for the guess you go closer to the other intersection point so we have the first curve the second curve and then the guess we go over to the other intersection point enter and we have two comma negative six so that's how we can find them by
• 21:00 - 21:30 graphing oh i missed a comma in there there that fixes it um generally speaking we're not going to solve by graphing because i mean that's a that's an estimate more than anything else but we can solve by substitution here we have y plus x equals 17 x squared plus y squared equals 169. we just need to solve for one of our variables x or y in the top one we're
• 21:30 - 22:00 going to plug it into the bottom one so if we solve for y we get y equals 17 minus x plug it in um and then we're going to simplify so first thing we have to do is do the 17 minus x squared and remember that is not 17 squared minus x squared don't be that guy we have 289 minus 34x plus x squared equals 169. it's a quadratic function let's move everything over to one side combine our like terms
• 22:00 - 22:30 we get 2x minus 34x plus 120 we can factor factor out the two we have x squared minus 17x plus 60. so now we need two things that multiply to give us 60 and add to give us negative 17 and people always think let's see 15 and 4 no 10 and 6 no remember that 60 is also 12 times 5. you have x minus 12 and x minus 5 equals zero so we can plug them in plug them into
• 22:30 - 23:00 the line because otherwise you when you square root you get plus or minus and a line and a circle which is what these are we'll have a maximum of two intersection points so x minus 12 equals zero so and x minus five equals zero so x equals 12 x equals five so plug them into y minus 17 we get 12 comma 5 and 5 comma 12.
• 23:00 - 23:30 we have something that looks like this there's a couple ways of doing it one you could subtract the 2 and plug that in you square and you get a quartic you get a y to the fourth or just solve for y squared get y squared equals 8x plus 16. and then plug it into y squared right we have y squared there to get x squared plus 8x plus 16 equals 100 and then again it's a quadratic so we're going to
• 23:30 - 24:00 subtract we're going to factor x plus 14 times x minus six so x equals negative 14 or six plug those in so we get negative 14 plus two equals 1 8 y squared which means negative 12 equals 1 8 y squared which means negative 96 equals y squared which you should see a problem happening here right no not that you can't square root 96 that it doesn't have to be a nice number
• 24:00 - 24:30 the fact that y squared equals a negative number that's problematic y squared is not going to equal a negative number so this is one that it doesn't work that happens so let's plug in the x equals six we get y squared equals 64. square root y equals plus or minus 8. so we have 6 comma 8 6 comma negative 8. and you can plug that into both of them and it'll work for both of them if you were to plug in negative 14
• 24:30 - 25:00 in no even in the x squared plus y squared equals 100 that wouldn't work out because that's a circle and negative 14 is outside of the circle so it's not going to hit that and you could also write it as 6 comma plus or minus 8 you just have to know that that is two solutions um when we get four solutions we could have like plus or minus six comma plus or minus eight that means all four possibilities
• 25:00 - 25:30 solving by elimination same basic idea except we're going to use elimination so we need one of our variables to be the same ex same coefficient except negative and then we'll add them together so here if we multiply that top equation by a negative one we'd have negative x squared minus y squared equals negative 20. we can add these together which would eliminate the y squared we get 3x squared equals 48 divide by 3 x squared equals 16 square root x equals plus or minus 4.
• 25:30 - 26:00 plug that in here it does not matter which one you plug it into because the plus or minus is going to be squared it will be positive so you'd have end up with 16 for either of them and subtract square root y equals plus or minus 2. so here we have all four possibilities we have 4 2 4 negative 2 negative 4 2 negative 4 negative 2 which can be written as plus
• 26:00 - 26:30 or minus 4 plus or minus 2. what plus or minus 4 comma plus or minus 2 does not mean is it does not mean 4 comma two negative four comma negative two it does not mean where we have four comma negative two negative four comma two it is all four of the possibilities um doing a similar one here we have an ellipse on top and a hyperbola on bottom
• 26:30 - 27:00 um we need the same value for each of them except once positive one's negative we could multiply that top one by three because notice the x squareds are here and here so rearranging them we have 27x squared minus 27x squared that's gonna cancel itself out so y squareds we end up with 21y squareds equals 189. we can divide 2y squared equals 9. so y
• 27:00 - 27:30 equals plus or minus 3. um this one just like the other one all both values are squared so when we plug it in we may as well plug in plug in y squared equals 9. like it's not going to matter if we plug in positive or negative 3. so we get 9x squared plus 5 times 9 equals 45. subtract 9x squared equals 0 which means x equals 0. so we have 0 comma plus or minus 3. this
• 27:30 - 28:00 one ends up only being two values um because of how it hits if we look we have a an ellipse and then we have a vertical hyperbola which is going to come and hit and come and hit sometimes it's fun to graph these just to see that it's like oh hey my answer actually did work that's what that thing looks like
• 28:00 - 28:30 and this could be written as zero comma plus or minus three so that is this third section looking at hyperbolas looking at solving systems of equations um i will see you in class but until then keep working problems keep asking questions and as always happy mapping