Hyperbolas - Conic Sections - By The Organic Chemistry Tutor
Transcript
00:0-1 | in this video , we're going to focus on hyperbole | |
00:02 | us . They may want to get a sheet of | |
00:04 | paper and a pen to write down some notes . | |
00:07 | So we have a horizontal hyperbole on the left and | |
00:10 | a vertical hyperbole on the right . The formula that | |
00:14 | corresponds to the horizontal hyperbole is x squared over a | |
00:19 | square minus Y squared over B squared . What do | |
00:23 | you think ? What you want ? This is the | |
00:25 | case when the hyperbole is centered at the origin , | |
00:30 | A . Is the distance between the center of the | |
00:34 | hyperbole and the vertex . So to get the two | |
00:40 | vortices , the coordinates are Plus or -10 for this | |
00:45 | type of hyperbole and the same is true for the | |
00:50 | other one . A . is the distance between the | |
00:53 | vertex and the center of the hyperbole . Now , | |
00:58 | the equation that corresponds to the vertical hyperba is as | |
01:02 | follows , it's y squared over a squared minus X | |
01:07 | squared over B squared , Which he goes one . | |
01:12 | So when a square is under X squared , it's | |
01:16 | going to open lesson , right ? Or when it's | |
01:19 | a positive sign in front of export , when there | |
01:24 | is a positive sign in front of y squared , | |
01:26 | it opens up and down and a square is always | |
01:30 | under the positive term . The squad is under the | |
01:35 | negative term . Now the coordinates for the vertex is | |
01:41 | for the vertical formula R zero comma plus or minus | |
01:45 | eight . The coordinates for the first I are plus | |
01:51 | or minus C comma zero on the left . And | |
01:54 | for that purple on the right , it's zero comma | |
01:56 | plus minus C . The hyperbole to I mean the | |
02:02 | full side is always located towards where the hyperbole opens | |
02:07 | up . Mhm and it's C units away from the | |
02:16 | center . Now the equation that correlates a B and | |
02:24 | C . For the hyperbole is c squared is equal | |
02:28 | to a squared plus b squared . In contrast for | |
02:35 | ellipses , the equation is different . C squared is | |
02:39 | equal to a squared minus b squared for an ellipse | |
02:44 | , A is always larger than B . But fiberglas | |
02:48 | A maybe larger than be and sometimes be maybe larger | |
02:51 | than A . So the relative sizes of those two | |
02:53 | are not important . They are important for ellipses but | |
02:56 | not for high prevalence . So make sure you're aware | |
02:58 | of that key difference and we'll go over plenty of | |
03:02 | examples so you'll see this in action . Now , | |
03:07 | the next thing you need to be familiar with is | |
03:08 | the transverse axis . The transverse axis is a line | |
03:13 | segment That connects the two verses together . So the | |
03:20 | left of the transverse axis is to A . Here | |
03:29 | we have a vertical transverse axis and here on the | |
03:32 | left is the horizontal transfers axis . The distance between | |
03:39 | the folks I or the two focal points is always | |
03:42 | going to be to see . But the length of | |
03:46 | the transverse axis will be to a Now let's talk | |
03:50 | about the equation of the ascent oats . For the | |
03:56 | horizontal hyperbole , it's Y is equal to plus a | |
03:59 | minus B Over eight times x . For the vertical | |
04:06 | hyperba it's Y is equal to plus or minus A | |
04:10 | over B times X . And this is true when | |
04:14 | the hyperbole to is centered at the origin , the | |
04:18 | equation is different if the center is not at the | |
04:21 | origin and we'll talk about that later in the video | |
04:24 | number one graph the hyperbole , identify the coordinates of | |
04:28 | the center vortices and fosse and then write the equations | |
04:32 | of the ascent oats . So what's the value of | |
04:36 | A squared and B squared in this problem , A | |
04:40 | squared is going to be under the variable that has | |
04:44 | a positive coefficient , but one of the negative sign | |
04:46 | is going to be associated with B squared . So | |
04:52 | in this problem a squared is the number under X | |
04:55 | squared . A square is four . A . It's | |
04:58 | going to be the square root of four , which | |
04:59 | is to B squared is equal to nine , And | |
05:04 | B is going to be the square root of nine | |
05:06 | , which is street . So with this information we | |
05:10 | can go ahead and draw a graph by the way | |
05:13 | , the center is going to be the origin for | |
05:17 | this particular problem . So A . Is two and | |
05:28 | it's under X squared . So we're gonna do is | |
05:31 | we're gonna travel to units to the right From the | |
05:35 | center and two units to the left . Let's put | |
05:37 | some marks on his graph 1st . So two units | |
05:54 | are right and then to units that left now be | |
05:58 | a street and it's under Y . So we're going | |
06:01 | to go up three units along the y axis from | |
06:04 | the center and then down three units . Then we're | |
06:08 | gonna do is we're gonna create a rectangle around these | |
06:12 | four points . The purpose of the rectangle is to | |
06:17 | help us to draw the asthma . Totes . The | |
06:21 | assam . Totes will be diagnose that will go right | |
06:27 | through the son of the box . So those are | |
06:42 | the two that to assume totes and what we have | |
06:47 | here represents the vertex is of the hyperbole . So | |
06:54 | whenever you have a positive X squared term , the | |
06:57 | hyperbole is gonna open towards the left and towards the | |
07:00 | right . So it's going to look something like this | |
07:08 | and then it's going to follow the asem toe . | |
07:19 | It's curved . Bye . The vortices . But once | |
07:23 | it follows the ascent , oh , it becomes almost | |
07:25 | linear . So that's how we can graph the hyperbole | |
07:35 | to to the best of our ability . So it's | |
07:39 | initially curve and then it becomes linear as it approaches | |
07:41 | the ascent oats . Now let's write the coordinates of | |
07:47 | the vortices . So the vertex is is going to | |
07:50 | be Plus or -10 for this type of graph And | |
07:55 | a is too . So it's plus or -2 common | |
07:58 | zero . So this one here is that two comma | |
08:02 | zero and this one here Is -20 . So those | |
08:10 | are the coordinates of the vortices . Now , let's | |
08:13 | focus on the coordinates of the full side . This | |
08:16 | is going to be plus or -10 . So we | |
08:20 | need to calculate the value of C and we could | |
08:23 | use this formula to achieve it . C squared is | |
08:26 | equal to a squared plus B squared , A squared | |
08:30 | is four , B squared is nine . So c | |
08:33 | squared is 13 which means see Is equal to the | |
08:37 | square root of 13 . So the coordinates of the | |
08:43 | folks i it's going to be plus or minus route | |
08:47 | 13 comma zero . The square root of 13 as | |
08:56 | a decimal value Is approximately 3.6 . So we have | |
09:02 | a focus there and the focus here . So those | |
09:05 | are the two focal points of the hyperbole . As | |
09:10 | you can see , the hyperbole always opens towards the | |
09:13 | focal points . Now let's write the equations of the | |
09:18 | asem totes . Now when you have an asem toe | |
09:22 | centered at the origin and if it opens to the | |
09:25 | right into the left , the equation is going to | |
09:28 | be why is equal to plus or minus be over | |
09:32 | A times X . This value in front of X | |
09:37 | is the slope . Its rise over run going from | |
09:43 | the origin to this point we have a rise of | |
09:47 | three . That's our B . Value and a run | |
09:49 | of A . So the slope is rise over run | |
09:52 | B over A . So that can help you determine | |
09:56 | if it's been over A or ever be if you | |
09:59 | forget so be history . A . Is too so | |
10:09 | why is equal to plus or minus 3/2 X . | |
10:14 | Now there's two equations for the two aspects tools that | |
10:16 | we have here , the one on the right . | |
10:19 | This is going to be why is equal to positive | |
10:23 | 3/2 x . As for the one on the left | |
10:27 | . This is why is equal to negative 3/2 X | |
10:34 | . And it makes sense because this line here , | |
10:37 | the slope is positive because it's increasing as you move | |
10:40 | to the right . As for the other line , | |
10:43 | the slope is negative because it's decreasing as you move | |
10:46 | to the right . So those are the two equations | |
10:48 | that correspond to the to ask tools that we have | |
10:51 | in this graph and the ascent filter linear . Which | |
10:55 | means that the hyperbole to becomes linear as X increases | |
11:00 | near the vertex is it's kind of curved but as | |
11:02 | X gets larger and it approaches the assent to it | |
11:05 | becomes it behaves like a linear function . Now let's | |
11:09 | work on a similar problem . So we're going to | |
11:12 | do the same thing . We're going to graph the | |
11:13 | hyperbole to identify the coordinates of the center of vortices | |
11:16 | and the folks I and then we're going to write | |
11:18 | the equations of the assam totes . Right now , | |
11:23 | this hyper Bella is not in standard form . We | |
11:26 | need this number to be a one . So what | |
11:29 | we're gonna do is we're gonna divide everything By 144 | |
11:37 | . 144 divided by nine is 16 . So we're | |
11:42 | going to get y squared over 16 . 1:44 divided | |
11:47 | by 16 is nine . So we're going to get | |
11:49 | X squared overnight And 144 divided by itself is one | |
11:57 | . So I'm just going to rewrite this here . | |
12:04 | Mhm . So now that our equation is in standard | |
12:09 | form , we can go ahead and find the details | |
12:12 | that we need . So what's a squared and what's | |
12:15 | B squared in this equation , A square is always | |
12:18 | going to be the one that has the positive coefficient | |
12:22 | . So a squared is going to be 16 . | |
12:27 | That means that A Is the square root of 16 | |
12:30 | , which is four . B squared is nine . | |
12:34 | B is going to be the square root of nine | |
12:36 | , which is three . So now let's calculate C | |
12:40 | C squared is equal to a squared plus B squared | |
12:44 | , A squared is 16 , b squared is nine | |
12:48 | , 16 plus nine is 25 And the square root | |
12:52 | of 25 is five . Now let's go ahead and | |
12:57 | draw a rough sketch of this graph . Now , | |
13:03 | since y squared is listed first , this is going | |
13:07 | to be a vertical hyperbole . So we're gonna travel | |
13:11 | a units along the Y axis since a squared is | |
13:16 | under Y squared and then a units down from the | |
13:20 | center . So just like before the center it's gonna | |
13:23 | be the origin , It's 0,000 , B squared is | |
13:33 | under X squared . And since vehstree we're going to | |
13:36 | travel three units to the left and to the right | |
13:40 | along the X axis . Now let's create our rectangle | |
13:57 | and then let's draw the assam totes which is going | |
14:01 | to go from one side , a direct tangle through | |
14:05 | the center to the other side . Now the coordinates | |
14:20 | of the vertex is it's gonna be zero plus or | |
14:23 | minus A And a . is four , so it's | |
14:26 | zero plus or -4 . Here's the first vertex at | |
14:30 | 0:04 And here's the second vertex at 0 -4 . | |
14:35 | And so the graph , it's gonna look like this | |
14:46 | . So that's how we can graph this particular hyperbole | |
14:51 | . So now that we have the coordinates of the | |
14:52 | vertex is and the center , we need to find | |
14:56 | the coordinates of the full side . So it's going | |
14:59 | to be zero comma plus or minus C And C | |
15:03 | is five . So at 05 . Yeah . And | |
15:09 | that 0 -5 , we have two focal points . | |
15:19 | So that's the coordinates for the full side . Now | |
15:23 | , the last thing we need to do is write | |
15:24 | the equations of the ascent oats . So it's gonna | |
15:27 | be y is equal to plus or minus the slope | |
15:31 | times X . The slope is going to be rise | |
15:34 | over run . So the rise is a units the | |
15:40 | run is being its survival run . It's going to | |
15:43 | be a over B A . Is four B.S Stream | |
15:55 | . And so it's going to be Y . Is | |
15:57 | equal to plus or minus 4/3 X . So that's | |
16:02 | how we can graft the hyperbole to identify the coordinates | |
16:05 | of the center advertises in full sight . And that's | |
16:07 | how we can write the equations of the ascent oats | |
16:11 | for this particular equation . Now , let's write down | |
16:15 | some more notes . So let's focus on The Horizontal | |
16:20 | Hyperbole 1st . When this hyperbole is centered at the | |
16:30 | origin , the equation that correlates to it is X | |
16:34 | squared over a squared minus Y squared over B squared | |
16:38 | . But what happens if the center is not at | |
16:41 | the origin ? Let's say it's at some point H | |
16:44 | com McCain the equation changes to this , it becomes | |
16:48 | x minus h squared over a squared minus y minus | |
16:55 | k squared over B squared is equal to one . | |
16:59 | So you may want to write some notes down . | |
17:01 | Even if you don't know how to apply it , | |
17:02 | you may just want to drop these down and then | |
17:04 | use it later . For reference . Now the next | |
17:08 | thing that we mentioned is that the vertex is for | |
17:12 | horizontal . Hyperbole is centered at the origin Is Plus | |
17:16 | or -10 . When it's not centered at the origin | |
17:21 | is going to change . All we need to do | |
17:23 | is at the center to this to those coordinates . | |
17:30 | So we're gonna add H two plus or minus K | |
17:33 | . I mean plus minus a rather . So it's | |
17:35 | H . Plus or minus A . And then we're | |
17:37 | gonna add zero with K . So it just becomes | |
17:41 | K . So that's how you can right ? The | |
17:43 | new Verdecia simply at the center to the vertex is | |
17:47 | and then you get the the new vortices . When | |
17:51 | the center is not at the origin , the coordinates | |
17:55 | of the folks i is plus or minus C comma | |
17:58 | zero . So when it's shifted and the center is | |
18:02 | not not at the origin simply add H and K | |
18:05 | to this . So it becomes a church plus or | |
18:10 | minus C . Calm . Okay , now the equation | |
18:17 | of the ascent , oats is plus or minus be | |
18:24 | over a times X . When the graph is shifted | |
18:28 | this becomes y minus K is equal to plus or | |
18:31 | minus B over a times x minus age . So | |
18:37 | that's how the equations will shift when the center moves | |
18:40 | from the origin to some point . H comma Kane | |
18:46 | . Now , what if we have the vertical ascent | |
18:47 | on how well the equations change . So first let's | |
18:53 | write the equations when it's centered at the origin . | |
18:58 | So for this one it's going to be y squared | |
19:00 | over a squared minus x squared over b squared . | |
19:05 | And when the center shifts from the origin to h | |
19:07 | comma k , the equation will be adjusted . So | |
19:11 | it's gonna be why minus K squared over a squared | |
19:15 | minus x minus h squared over B squared which equals | |
19:20 | one . The very disease will be zero comma plus | |
19:26 | or minus A . The new vertex is all we're | |
19:28 | gonna do is at the center to this . So | |
19:31 | it becomes h comma K plus or minus A . | |
19:39 | The full side will be zero plus or minus C | |
19:42 | . The new coordinates of the folks , I will | |
19:44 | be H comma K plus or minus C . The | |
19:51 | equation for the ascent oats will be plus or minus | |
19:55 | a over B . X . But the new equation | |
19:58 | becomes y minus K . Is equal to plus or | |
20:01 | minus a over B x minus h . So feel | |
20:06 | free to write those equations . This equation remains the | |
20:09 | same . C squared is equal to a squared over | |
20:12 | b squared . In all cases , number three graph | |
20:18 | the hyperbole . So we're going to follow the same | |
20:20 | instructions that we did for the first two problems . | |
20:23 | But this time we can see that it's not centered | |
20:25 | at the origin . So let's begin by finding the | |
20:28 | center . Here we have ex monastery . We're going | |
20:32 | to change -3 . It's a positive three . And | |
20:35 | here we have y plus two . We're going to | |
20:37 | change positive to to negative through . So the coordinates | |
20:41 | of the center , Our 3 -2 . A squared | |
20:47 | is typically the first number that we see here . | |
20:50 | A squared is four . So a is going to | |
20:52 | be too B squared is nine . So be it's | |
20:56 | gonna be three . Next we need to calculate C | |
21:03 | . A squared is four . B squared is nine | |
21:06 | , four Plus 9 is 13 . So see it's | |
21:09 | going to be the square root of 13 . Now | |
21:14 | let's go ahead and graph the hyperba . So let's | |
21:41 | begin by plotting the center which is that three negative | |
21:44 | two . Now A is to A is associated with | |
21:49 | the X . Variable . So we're going to travel | |
21:51 | to units to the right parallel to the X axis | |
21:55 | and two units to left be history B is B | |
22:00 | squared , was under Y squared . So it's associated | |
22:03 | with why ? So we're gonna go three units up | |
22:06 | parallel to the y axis and three units down from | |
22:09 | the center parallels the y axis . Now let's draw | |
22:14 | a rectangle and then let's draw the ascent oats . | |
22:44 | Now this particular hyperbole to will it open to the | |
22:48 | left and right ? Or will it open up and | |
22:50 | down ? Do we have a horizontal hyperbole ? A | |
22:53 | Or would you say we have a vertical hyperbole ? | |
22:57 | Now we have a positive sign in front of X | |
22:59 | squared . So this is going to open left and | |
23:01 | right . If y squared was positive it would open | |
23:05 | up and down . So the graph is going to | |
23:09 | look something like this . So these two points represent | |
23:20 | the vertex is of the hyperbole . And if you | |
23:23 | recall divert is is for a horizontal hyperbole to that | |
23:29 | is not centered at the origin . It's gonna be | |
23:32 | a church plus or minus a comma Kane H history | |
23:43 | A is to and Kay is negative two . So | |
23:51 | you can always look at the center to find H | |
23:52 | and K . So first we have three plus two | |
23:57 | which is five common negative too . So that's the | |
24:03 | vertex on the right side for this one it's gonna | |
24:07 | be three minus two comma negative 23 minus two is | |
24:10 | one . And so we get the point one negative | |
24:14 | two for this very text on the left . So | |
24:18 | those are the coordinates of the vertex is . Now | |
24:22 | let's find the coordinates of the full side . So | |
24:26 | it's gonna be a church plus or minus C . | |
24:29 | Can McCain hs three . See if the square root | |
24:35 | of 13 K is -2 . So this represents the | |
24:42 | coordinates of the folks I now to get the X | |
24:47 | value Of the folks I we have three plus the | |
24:50 | square to 13 . The square to 13 is 3.6 | |
24:53 | . So three plus 3.6 is 6.6 . So somewhere | |
24:58 | in this region we have to focus and then three | |
25:07 | minus Discord 13 , which is negative 130.6 . So | |
25:11 | somewhere in this region is the other focal point . | |
25:17 | So that's the location of the full site on the | |
25:20 | graph . The last thing we need to do is | |
25:22 | write the equations of the ascent oats . So it's | |
25:26 | going to be y minus K is equal to plus | |
25:31 | or minus . Now the rise is the units . | |
25:38 | The run is A . So the slope is going | |
25:40 | to be B over A and then it's x minus | |
25:45 | h . So Kay is -2 . Why minus negative | |
25:52 | two becomes y plus two . So what we have | |
25:57 | here is similar to the original equation . This is | |
26:00 | going to be equal to plus or minus B over | |
26:02 | a . B . History is too and then x | |
26:05 | minus h . So we have x minus tree . | |
26:12 | So this represents the equations of the to assume totes | |
26:17 | . Now let's work on another problem . So we're | |
26:19 | going to do what we did in the last problem | |
26:21 | . We're going to grab the hyperbole to identify the | |
26:24 | coordinates of the center , vortices and foresight . We're | |
26:28 | gonna write the equations of the asthma totes but this | |
26:30 | time we're also going to determine the domain and range | |
26:33 | of the graph . So what's a squared and what's | |
26:38 | B squared ? A squad is usually the number in | |
26:40 | front . So a square is going to be nine | |
26:45 | , which means A . Is a square to nine | |
26:48 | , which is street B squared is going to be | |
26:50 | the other 1 . 16 Be the square to 16 | |
26:54 | is four . Now let's calculate C C squared is | |
26:59 | a squared plus B squared , A squared is nine | |
27:04 | , B squared is 16 , nine plus 16 is | |
27:08 | 25 And the square root of 25 is five . | |
27:15 | Now , let's determine his center of this particular hyperbole | |
27:22 | . Here we see negative too . So the X | |
27:24 | coordinate of the center will be positive too . Here | |
27:27 | we see -1 . The y coordinate the center will | |
27:30 | be positive one . Now let's go ahead and plot | |
27:34 | it . So let's begin by plotting the center Which | |
28:07 | is at 2:01 . And then we see that a | |
28:11 | history A is under Y squared . So we're gonna | |
28:15 | go up three And then we're gonna go down three | |
28:26 | . Now bs four B . Is under the X | |
28:28 | . Variable to think of left and right when dealing | |
28:30 | with X . So we're gonna go right for from | |
28:33 | the center And then left four units . Now let's | |
28:38 | draw a rectangle . Next let's write out or draw | |
28:55 | the ascent oats . Let's do the same for the | |
29:03 | other one . I think I messed it up . | |
29:18 | Let's do this one more time . I know my | |
29:21 | graph is not drawn to scale but we'll make the | |
29:30 | best of that . So now the graph in part | |
29:35 | , so we know it's going to be curved at | |
29:39 | the the vertex . And then we need to follow | |
29:43 | the aceto . So let's try to draw an accurate | |
29:50 | graph . So that's as best as we can do | |
30:01 | for it . Now , as long as you draw | |
30:03 | up sketch , most teachers will give you , you | |
30:06 | know , the correct answer . But ideally you want | |
30:09 | to follow the aceto as you move away from the | |
30:13 | vertex is so that's the hyperbole that we can draw | |
30:17 | for this particular problem . Now we have the coordinates | |
30:22 | of the center . So now let's focus on devices | |
30:26 | . So for this particular type of graph where we | |
30:29 | have a vertical hyperbole to diversity is is going to | |
30:33 | be according to the verses will be h comma K | |
30:37 | plus or minus A . So H is two , | |
30:42 | K is one , so it's too common one plus | |
30:45 | or minus A . Which is street . So this | |
30:49 | point here , this vertex that's going to have an | |
30:52 | X . Value of two and then a Y . | |
30:54 | Value of one plus three . Which is for now | |
30:58 | the vertex below that will have the same X value | |
31:01 | but the Y value will be one . Monastery . | |
31:04 | So the point will be -2 2 -2 . So | |
31:09 | those are the coordinates of the vortices to come up | |
31:12 | for And to -2 . Now let's focus on finding | |
31:17 | the coordinates of the false I . So the coordinates | |
31:21 | will be h comma K Plus or -1 , H | |
31:31 | is two , K . is one And C is | |
31:34 | five . So the first one is going to be | |
31:39 | At 2:06 , which will be somewhere here . The | |
31:44 | next one will be too And then it's gonna be | |
31:47 | 1 -5 which is -4 . So I'm just gonna | |
31:51 | write that here . So this one is 2:06 And | |
31:54 | this will be 2 -4 . So those are the | |
31:58 | coordinates are the focus on . Now let's focus on | |
32:02 | the equations of the ascent , oats . The form | |
32:06 | that we need is why minus K . Is going | |
32:09 | to equal plus or minus the slope . And so | |
32:12 | we have a rise of a , a run of | |
32:16 | B . So it's gonna be a over B and | |
32:24 | then x minus h . So why minus K ? | |
32:27 | It's gonna be what we see here . That's why | |
32:31 | I -1 is equal to plus or minus AS . | |
32:35 | three , B is four and then x minus h | |
32:42 | is what we see here . So X -2 . | |
32:47 | So those are the equations of the assam totes . | |
32:52 | Now , the last thing we need to do is | |
32:53 | find the domain and range of the graph for a | |
32:58 | vertical hyperbole to the domain is always going to be | |
33:01 | our poll numbers . When you need to find a | |
33:03 | domain . Look at the X valleys , analyze it | |
33:06 | from left to right , the lowest x value will | |
33:09 | be negative infinity and the highest x value is positive | |
33:13 | , singing . So X can be anything along this | |
33:17 | curve . This curve is continuous . There are no | |
33:19 | jumps as you go from left to right . So | |
33:24 | the domain is our phone numbers . Now , the | |
33:26 | range is a different story . We need to look | |
33:28 | at the Y values , so view it from going | |
33:31 | from the bottom to the top , this can go | |
33:34 | all the way down to negative infinity . So the | |
33:37 | lowest Y value is negative infinity . As we go | |
33:41 | up This stops at AY Value of -2 . Now | |
33:48 | between a y value of -2 and four , there | |
33:52 | is no curve , there is no hyperbole . So | |
33:55 | we're going to pick it up back at four and | |
33:57 | then this goes up to infinity . So the range | |
34:03 | from bottom to top is negative infinity to -2 . | |
34:08 | Negative two is included because that's a vertex . So | |
34:11 | we're going to use a bracket union , We're going | |
34:13 | to start back up at four and continue to infinity | |
34:18 | . So that's how we can write the range us | |
34:22 | a hyperbole . And this is the domain for a | |
34:25 | vertical hyperbole . |
Summarizer
DESCRIPTION:
This conic sections video tutorial provides a basic introduction into hyperbolas. It explains how to graph hyperbolas and how to find the coordinates of the center, vertices, and foci. In addition, it explains how to write the equations of the asymptotes.
OVERVIEW:
Hyperbolas - Conic Sections is a free educational video by The Organic Chemistry Tutor.
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