Finding The Focus and Directrix of a Parabola - Conic Sections - By The Organic Chemistry Tutor
Transcript
| 00:00 | in this video , we're going to focus on problems | |
| 00:04 | . So let's talk about some equations that you need | |
| 00:06 | to know . So hopefully you have a sheet of | |
| 00:08 | paper with you and a pen and write down some | |
| 00:09 | notes . So for the problem on the left , | |
| 00:13 | this corresponds to the equation Y squared is equal to | |
| 00:17 | four p . x . And you would use this | |
| 00:21 | if the vertex of the problem is the origin . | |
| 00:27 | For the equation on a right , It's x squared | |
| 00:31 | is equal to four py . With the vertex being | |
| 00:36 | at the origin as well . The focus is to | |
| 00:42 | the right where the problem opens towards . So for | |
| 00:47 | this problem it's somewhere in this region . P is | |
| 00:52 | the distance between the vertex and the focus . Now | |
| 00:58 | , if you go pee units in the other direction | |
| 01:01 | , you're going to get something called the rhetoric . | |
| 01:08 | The rhetoric is P units away from the vertex for | |
| 01:13 | this particular graph , it's going to be X is | |
| 01:16 | equal to negative P . Now , when P is | |
| 01:21 | positive , the graph is going to open towards the | |
| 01:25 | right . When P is negative , it's gonna open | |
| 01:29 | towards the left . For this graph , the focus | |
| 01:35 | would be somewhere in this region . It's going to | |
| 01:39 | be P units away from the vertex and then P | |
| 01:44 | units below that . We have . The rhetoric sucks | |
| 01:48 | . In this case we have a horizontal metrics . | |
| 01:51 | That's the equation will be Y is equal to negative | |
| 01:54 | P as opposed to X is equal to negative P | |
| 01:59 | . Now going from the focus to the curve , | |
| 02:03 | That's a distance of two p . And going in | |
| 02:09 | the other direction Is a distance of two p . | |
| 02:16 | So the segment that connects these two points on the | |
| 02:20 | curve and that passes through the focal point . That | |
| 02:23 | segment is known as the lattice rectum . The lattice | |
| 02:30 | rectum Has a left of four p . It's basically | |
| 02:36 | the focal diameter of the problem . Now , when | |
| 02:41 | P is positive , the problem is going to open | |
| 02:45 | upward if you have this equation and when P is | |
| 02:48 | negative it's going to open . And that word , | |
| 02:54 | So that's a simple introduction into problems . But let's | |
| 02:57 | begin working on some problems by the way , for | |
| 03:00 | those of you who want access to the full version | |
| 03:02 | of this video , feel free to take a look | |
| 03:05 | at the links in the description section below this video | |
| 03:09 | . So let's go ahead and continue graph the equation | |
| 03:12 | shown below . So this equation is in the form | |
| 03:17 | X squared Is equal to four py with the vertex | |
| 03:23 | centered at the origin . There's no asian came this | |
| 03:27 | equation which we'll talk later about that in the video | |
| 03:32 | . Now the first thing we need to do is | |
| 03:33 | calculate P four P . Is the number in front | |
| 03:37 | of wine . So we're gonna set four p equal | |
| 03:41 | to eight , Dividing both sides by four P . | |
| 03:45 | is going to be a divided by four . Which | |
| 03:47 | is to . Now once we have our P value | |
| 03:50 | we can go ahead and graph the equation . So | |
| 04:02 | first let's plot P . The vertex is at the | |
| 04:06 | origin and going to units up we get the focus | |
| 04:13 | which is here . So the focus is at 0:02 | |
| 04:21 | . Now once you have the focus , what you | |
| 04:24 | want to do is you want to go to p | |
| 04:26 | units to the right and to P units LF If | |
| 04:29 | he is to to PS four . So we're gonna | |
| 04:33 | travel four units to the right and four units to | |
| 04:36 | the left to get to additional points . And then | |
| 04:44 | we can draw a rough sketch of the hyperbole . | |
| 04:53 | So keep in mind this here is to pee and | |
| 04:56 | this part is to pee . Now let's write the | |
| 05:02 | equation of the metrics . So we're going to go | |
| 05:05 | pee units below the vertex . And then we're gonna | |
| 05:09 | draw horizontal dash line . So that line is that | |
| 05:15 | why is equal to -2 ? And that's the equation | |
| 05:18 | of the metrics . So that's it for this problem | |
| 05:21 | . That's how we can graph this particular problem . | |
| 05:25 | Number two , graph the parabolic equation shown below . | |
| 05:30 | Find the coordinates of the focus and write the equation | |
| 05:33 | of the direct tricks and determine the left of the | |
| 05:36 | lattice rectum . So our general formula is going to | |
| 05:40 | be y squared is equal to four PX . So | |
| 05:44 | let's begin by 7 . 4 p . equal to | |
| 05:47 | the number in front of us which is to So | |
| 05:52 | four p . is equal to two , Dividing both | |
| 05:55 | sides by four . We get P is to over | |
| 05:58 | four Which reduces to 1/2 . So now that we | |
| 06:04 | have the value of P . And we know that | |
| 06:06 | the vertex is centered at the origin . Let's go | |
| 06:11 | ahead and graphic . So for this equation , the | |
| 06:17 | graph is going to open towards the right since P | |
| 06:21 | is positive , the vertex is at the origin . | |
| 06:27 | Let's put some points here since he is mom gonna | |
| 06:31 | space out each point Now P is one half , | |
| 06:41 | So .5 units to the right of the vertex will | |
| 06:46 | be the focus and I'm going to travel up two | |
| 06:54 | p units and down to P units . If he | |
| 06:57 | is a half two , P is going to be | |
| 06:59 | one . So the next point will be here and | |
| 07:04 | the other point will be here . So the graphic | |
| 07:07 | , it's going to look something like this . Let's | |
| 07:09 | see if I can draw much smoother graph , We'll | |
| 07:19 | go with that . So the coordinates of the focus | |
| 07:24 | is the X value is one half and the y | |
| 07:28 | value is zero . Now going 1/2 units in the | |
| 07:32 | other direction , we're going to get the equation for | |
| 07:37 | the metrics and that's that X equals negative P or | |
| 07:43 | X equals negative one half . So now we have | |
| 07:47 | the equation of the metrics and we have the coordinates | |
| 07:51 | of the focus . The last thing we need to | |
| 07:54 | do is determine the left of the latest rectum or | |
| 07:57 | the latest rectum and it's basically the focal diameter connecting | |
| 08:04 | these two points on the curve . So we know | |
| 08:06 | this part is to p this part is to pee | |
| 08:09 | . Thus the left Of the lattice rectum is always | |
| 08:13 | going to be four p . So p is a | |
| 08:16 | half . four times a half is too . So | |
| 08:20 | the answer to the last part of the problem is | |
| 08:22 | two units long number three match each equation to one | |
| 08:27 | of the graphs shown below . So let's start with | |
| 08:31 | this one , X squared is equal to four Y | |
| 08:34 | . When you have Y and x squared , that's | |
| 08:37 | the graph that's gonna open up or down now . | |
| 08:39 | There are no negative signs here . So this one | |
| 08:42 | is going to open up , so therefore this corresponds | |
| 08:45 | to answer choice A . The next one Why squared | |
| 08:49 | is equal to four y . When you have Y | |
| 08:52 | squared and X to the first power . This is | |
| 08:54 | going to open either to the right or two left | |
| 08:57 | , but since there's a negative sign , it's going | |
| 08:58 | to open to the right . So this is B | |
| 09:01 | . Now for C Y squared equals negative four Y | |
| 09:04 | . It's opposite to this one . So if this | |
| 09:07 | one opens to the right , why squared equals negative | |
| 09:10 | four , Y is going to open to the left | |
| 09:13 | , the last one , D x squared equals negative | |
| 09:16 | four , Y . Is opposite to A . So | |
| 09:19 | this one is going to open downward . So make | |
| 09:22 | sure you understand how to associate the correct orientation of | |
| 09:26 | the problem with the appropriate equation . Because you need | |
| 09:30 | to know this if you're going to study for tests | |
| 09:32 | and this is gonna be very important whenever you need | |
| 09:34 | to graph a parabola , especially the heart of versions | |
| 09:39 | . So just to review when X is squared and | |
| 09:41 | why is not it's going to open either up or | |
| 09:45 | down . So for this woman it opens up , | |
| 09:47 | we have positive for Y . When it opens down | |
| 09:51 | , there's gonna be a negative sign . So this | |
| 09:53 | is X squared is equal to negative four Y . | |
| 09:57 | Now when Y is squared and X is not , | |
| 09:59 | it's gonna open either to the right or to the | |
| 10:01 | left . When P is positive , it's going to | |
| 10:04 | open to the right . When P is negative , | |
| 10:07 | it's gonna open to the left . So make sure | |
| 10:11 | you understand us number four right . The standard form | |
| 10:15 | of the equation for the problem with the given conditions | |
| 10:18 | . So we're given the coordinates of the focus and | |
| 10:21 | the equation of the metrics . So let's go ahead | |
| 10:25 | and plot the information that we have here . So | |
| 10:27 | let's start with the focus . So the focus is | |
| 10:36 | at -30 . So there it is And then we | |
| 10:41 | have the rhetoric six at X equals positive three . | |
| 10:51 | Well , we need to find is the vertex the | |
| 10:54 | vertex of the problem is going to be the midpoint | |
| 10:56 | between the focus and the rhetoric . The focus . | |
| 10:59 | Is that positive 3 ? The rhetoric is that -3 | |
| 11:03 | . So when you average those two numbers , you'll | |
| 11:06 | see that the vertex is going to be at the | |
| 11:07 | origin 00 So , to draw a rough sketch of | |
| 11:13 | the graph , we know that it's going to open | |
| 11:16 | toward the focus away from the metrics . So this | |
| 11:19 | graph is going to open towards the left . Now | |
| 11:24 | , in order to write the equation , we need | |
| 11:26 | to calculate P P . Is the distance between the | |
| 11:31 | vertex and the focus and also between the focus and | |
| 11:34 | the direct tricks . In both cases we could see | |
| 11:37 | that P is three units long Now it's P positive | |
| 11:45 | three or -3 . The fact that the hyper but | |
| 11:49 | I mean that's hyperbole but the parabola opens towards the | |
| 11:52 | left towards the focus means that he is negative , | |
| 11:55 | so P is negative three . In this case , | |
| 12:00 | This part here is going to be two p And | |
| 12:04 | that part is going to be two p below the | |
| 12:06 | focus to PS six . So if you want to | |
| 12:09 | draw an accurate sketch , This point here should be | |
| 12:14 | at a y value of six , And this point | |
| 12:17 | here should be at AY . Value of -6 . | |
| 12:21 | Now my sketches not drawn to scale , but those | |
| 12:24 | would be the points that is part of the lactose | |
| 12:30 | rectum . So now that we have our P . | |
| 12:33 | Value , we can now write the equation . So | |
| 12:36 | what's the general form of the equation for parabola ? | |
| 12:40 | Where the vertex is at the origin but it opens | |
| 12:42 | to the left . This is going to be Why | |
| 12:46 | squared is equal to four p . x . For | |
| 12:50 | horizontal problem that opens to the left or to the | |
| 12:53 | right . Now all we need to do is plug | |
| 12:55 | in P . So P is -3 , four times | |
| 13:00 | -3 is -12 . So this here is the final | |
| 13:04 | answer . This is the standard form of the equation | |
| 13:08 | for the problem With a focus of native to become | |
| 13:11 | a zero . And in the rhetoric six of x | |
| 13:15 | equals positive three . So the answer is why squared | |
| 13:18 | is equal to negative 12 x . Number five identified | |
| 13:23 | the coordinates of the vertex and focus Write the equation | |
| 13:27 | of the rhetoric six graph the proble calculate the left | |
| 13:31 | of the lattice rectum and determined the domain and range | |
| 13:34 | of the function . So let's begin . So this | |
| 13:38 | equation is in the form Why squared is equal to | |
| 13:42 | four p . x . Now we know that equation | |
| 13:50 | has its vertex at the origin and it opens to | |
| 13:53 | the right if P is positive and we have a | |
| 13:57 | direct tricks here . P units away from the vertex | |
| 14:05 | . Now we do have some numbers associated with X | |
| 14:09 | . And Y . So this particular problem has been | |
| 14:12 | shifted away from the origin . The Vertex was 000 | |
| 14:18 | . Now it has been shifted to some point H | |
| 14:23 | comma K . When it's been shifted , this equation | |
| 14:26 | changes to this . It becomes why minus K . | |
| 14:31 | Squared Is equal to four p x minus H . | |
| 14:36 | Where H and K are the coordinates of the vertex | |
| 14:41 | . Now , the focus , which was here , | |
| 14:43 | initially it initially had the point P comma zero . | |
| 14:53 | But now once you add the new vertex to it | |
| 14:55 | , once you add H comma K , the new | |
| 14:57 | focus becomes H plus P . Com Okay , as | |
| 15:05 | for the equation of the direct tricks , it was | |
| 15:07 | X equals negative P . But once you add H | |
| 15:09 | to it , it becomes X is equal to h | |
| 15:13 | minus P . So let's go ahead and work on | |
| 15:17 | this problem . Hopefully you wrote those down , but | |
| 15:24 | as long as you understand that you can get everything | |
| 15:26 | that you need . But let's write this equation , | |
| 15:29 | y minus k squared is equal to four P x | |
| 15:32 | minus H . And let's find the vertex . So | |
| 15:37 | the very text is going to be hk McCain , | |
| 15:41 | so looking at the number associated with X , it's | |
| 15:43 | a plus to to find a church , simply reverse | |
| 15:46 | the sign . So this is going to be negative | |
| 15:48 | too . And the number in front of why is | |
| 15:51 | -3 . So reverse it to positive three . And | |
| 15:54 | now we have hnk so H is negative two K | |
| 15:58 | A stream . Now the next thing we need to | |
| 16:04 | do is find a focus and a vertex . But | |
| 16:07 | let's start with the graph . So the vertex , | |
| 16:34 | is that negative ? To comment three , which is | |
| 16:36 | here . And now let's calculate P . So four | |
| 16:44 | is equal to four . p Dividing both sides by | |
| 16:49 | four . We get P is equal to one . | |
| 16:53 | So now that we know what P is and P | |
| 16:56 | is one . So we know that this is going | |
| 16:58 | to open towards the right . If we travel one | |
| 17:01 | units to right , we'll get the focus and then | |
| 17:07 | we're going to go up to pee and down to | |
| 17:10 | pee . So if P is 12 P is two | |
| 17:14 | . So that will take us to this point at | |
| 17:16 | that point . And now we can graph . Let | |
| 17:20 | me use a different color . So this is how | |
| 17:25 | probable is going to look like . Now let's travel | |
| 17:31 | p units to the left . P . Is one | |
| 17:35 | . So the rhetoric will be one unit away from | |
| 17:38 | the vertex . So we can clearly see that the | |
| 17:45 | equation of the metrics will be X is equal to | |
| 17:48 | negative three . Any time the metrics is vertical it's | |
| 17:51 | gonna be X . Is equal to a number . | |
| 17:53 | If it's horizontal it's why is equal to a number | |
| 17:57 | . The formula for calculating the matrix is H minus | |
| 18:00 | P . For this type of shape . But if | |
| 18:02 | you don't want to memorize the formula , you could | |
| 18:04 | just look at the graph and see what the answer | |
| 18:05 | is , H is -2 , p Is positive one | |
| 18:11 | . It's a -2 -1 . You get -3 . | |
| 18:14 | For those of you who want to see how to | |
| 18:16 | use that equation . Now the focus to find accordance | |
| 18:21 | to the focus , we could just look at the | |
| 18:22 | graph . We can see it has an x . | |
| 18:24 | value of negative one And the Y Value of three | |
| 18:28 | . Using the formula it's H plus P . Comic | |
| 18:32 | A . For this type of horizontal problem , H | |
| 18:40 | is -2 . P is one . K . history | |
| 18:44 | , so negative two plus one is negative one . | |
| 18:47 | So we get that the focus Is -1 , commentary | |
| 18:54 | . So right now we have the coordinates of the | |
| 18:55 | focus . We have the equation of the metrics . | |
| 19:02 | We also have the coordinates of the vertex and we | |
| 19:07 | have the graph of the problem . The next thing | |
| 19:10 | we need to do is find the left of the | |
| 19:11 | lattice rectum . So that's gonna be the distance between | |
| 19:17 | those two points , The left of the Lattice Rectum | |
| 19:22 | , as we've considered before , it's always going to | |
| 19:24 | be equal to four p . And since P is | |
| 19:26 | one , this is going to equal for the last | |
| 19:31 | thing we need to do is determine the domain and | |
| 19:32 | range of the function . So for the domain we're | |
| 19:36 | looking at the X values from left to right . | |
| 19:39 | The lowest X value is the X coordinate of the | |
| 19:42 | vertex , which is negative two . Since -2 is | |
| 19:47 | part of the graph . We included with a bracket | |
| 19:50 | . Now this is going to go all the way | |
| 19:51 | to the right , all the way to positive infinity | |
| 19:54 | . So the domain is from negative to to infinity | |
| 19:57 | . Now , for the range we're going to focus | |
| 19:59 | on the Y values from the bottom to the top | |
| 20:03 | . This graph will keep on going forever . It's | |
| 20:05 | going to go to the right and it's going to | |
| 20:06 | go down . So it's gonna go down all the | |
| 20:09 | way to negative infinity . And you can follow all | |
| 20:12 | of the Y values along this curve as it goes | |
| 20:15 | up to positive infinity . So for horizontal problem that | |
| 20:21 | opens to the left or to the right , the | |
| 20:24 | range is always going to be our poll numbers . | |
| 20:28 | If you have a vertical problem that opens up and | |
| 20:30 | down then the domain will be are real numbers negative | |
| 20:33 | infinity to positive infinity . Number six , identify the | |
| 20:38 | coordinates of the vertex and focus and basically do everything | |
| 20:42 | that we did in the previous problem . So the | |
| 20:46 | standard form of the problem , this problem is going | |
| 20:50 | to be x minus h squared Is equal to four | |
| 20:53 | p times y -K . And the vertex is going | |
| 20:57 | to be at hk . So let's begin by finding | |
| 20:59 | the vertex first Here we see -3 . So we're | |
| 21:03 | gonna have positive three for X for why we see | |
| 21:06 | negative too . So we're going to change the positive | |
| 21:08 | too . So that's the coordinates of the vertex for | |
| 21:12 | this particular problem . So HS three K is too | |
| 21:18 | . Now let's calculate P -8 is equal to four | |
| 21:22 | . p . Let's divide both sides by four . | |
| 21:29 | And so P is gonna be negative 8/4 . Which | |
| 21:32 | is to Now let's go ahead and sketch a grass | |
| 21:41 | to this problem . So let's begin by plotting a | |
| 22:07 | vertex which is at three common to now . What | |
| 22:14 | direction will the graph open ? Will open to the | |
| 22:17 | right ? Will open to the left . Will it | |
| 22:21 | open up or will it open down ? What would | |
| 22:26 | you say ? The first thing we need to pay | |
| 22:32 | attention to is which variables are squared and which ones | |
| 22:36 | are not X is squared . Why is not ? | |
| 22:41 | So whenever you have Y equals X squared . This | |
| 22:43 | is a problem that opens up or down . And | |
| 22:47 | since P is negative , this is a problem that's | |
| 22:50 | going to open in the downward direction , So P | |
| 22:56 | is -2 . We're going to go down two units | |
| 23:01 | and this will give us our focus which is here | |
| 23:10 | , so we can see that the coordinates of the | |
| 23:12 | focus Is at 30 . The form , let's calculate | |
| 23:18 | the focus for this type of problem that opens up | |
| 23:22 | or down . It's gonna be h comma K plus | |
| 23:28 | P H . And this example of ST K is | |
| 23:36 | two and P is -2 , two plus negative two | |
| 23:42 | is zero . So we get the 0.3 comma zero | |
| 23:49 | . So now that we have the coordinates of the | |
| 23:51 | focus , let's find the other points that we need | |
| 23:57 | in order to graph this problem . So we're going | |
| 24:01 | to travel two P units to the right and two | |
| 24:05 | p units to left P is -2 . So two | |
| 24:09 | P is gonna be negative form . But we're gonna | |
| 24:11 | use the absolute value of that . So the absolute | |
| 24:14 | value of two PS 4 . So we're gonna travel | |
| 24:17 | four units to the right from the focus And then | |
| 24:22 | four units to the left . And now we can | |
| 24:26 | sketch our problem . So now let's go up p | |
| 24:35 | units . This will give us the equation of the | |
| 24:40 | metrics . So here the Y values to here it's | |
| 24:46 | gonna be four . Thus the equation of the rhetoric | |
| 24:49 | will be why is equal to positive . For when | |
| 24:54 | dealing with a vertical problem . The equation that you | |
| 24:57 | need to calculate the rhetoric is this why is equal | |
| 25:00 | to k minus p . K . Is too P | |
| 25:06 | is -2 . So you get to minus negative too | |
| 25:10 | , Which is the same as 2-plus 2 . And | |
| 25:13 | so you get why is equal to four . Now | |
| 25:21 | let's determine the length of the lattice rectum . So | |
| 25:25 | the latest rectum is the focal diameter That connects two | |
| 25:29 | points on the curve and passes through the focus . | |
| 25:33 | So you know , this part is to pee . | |
| 25:35 | The other part is to pee . So that gives | |
| 25:38 | us a total length of four p . Now he | |
| 25:49 | is negative . And the left of that segment , | |
| 25:51 | we're just going to assign a positive value to it | |
| 25:54 | . So technically we need to say that the length | |
| 25:56 | of the lattice rectum is the absolute value of four | |
| 25:59 | p . So that's the absolute value of four times | |
| 26:03 | negative two . Which will give us eight . So | |
| 26:06 | the length of the lattice rectum in this example is | |
| 26:09 | eight units long . Now let's focus on the domain | |
| 26:12 | and range . Starting with the domain . For a | |
| 26:15 | vertical problem , it's going to be horrible numbers if | |
| 26:17 | you analyze it from left to right , the lowest | |
| 26:20 | X value is negative infinity . The highest X value | |
| 26:23 | is infinity , and X could be any number in | |
| 26:26 | between those two extremes . So for any vertical problem | |
| 26:31 | , the domain is always all real numbers . The | |
| 26:34 | range is different though , so for the range will | |
| 26:36 | analyze it from the bottom to the top . The | |
| 26:38 | lowest Y value . This can go down forever . | |
| 26:41 | So the lowest Y value is negative infinity . A | |
| 26:45 | vertical probably like this one has a maximum . If | |
| 26:47 | it opens downward it has a minimum . If it | |
| 26:50 | opens upward , the maximum is basically the vertex . | |
| 26:55 | The white corner of the vertex is too . So | |
| 26:58 | the range is going to be from negative infinity up | |
| 27:02 | to the Y coordinate the vertex , which is to | |
| 27:08 | . So that's how you can determine the domain and | |
| 27:09 | range for a parabola . Number seven , write the | |
| 27:15 | equation of the problem in standard form , identify the | |
| 27:19 | coordinates of the vertex and focus , write the equation | |
| 27:22 | of the rhetoric and graph the problem . So we | |
| 27:27 | have the equation in non standard form , so to | |
| 27:30 | speak . We need to put it in standard form | |
| 27:32 | . How can we do this ? Well , we're | |
| 27:35 | going to have to use a technique called completing the | |
| 27:37 | square . But first we need to know what type | |
| 27:40 | of problem we're dealing with , notice which variable is | |
| 27:44 | squared and which one is not X is squared , | |
| 27:48 | Why is not ? So therefore we need to put | |
| 27:51 | this equation in this format X minus h squared is | |
| 27:55 | equal . It's a four p times y minus K | |
| 28:02 | . So how can we turn this equation into something | |
| 28:05 | that looks like that ? Well , we could see | |
| 28:08 | that the X variables are on the left . Why | |
| 28:11 | is on the right ? So that gives us some | |
| 28:13 | guidance in terms of what we need to do here | |
| 28:17 | . Sometimes this software acts up . What we're gonna | |
| 28:21 | do is we're going to keep the X variables on | |
| 28:24 | the left side so we can mirror what we have | |
| 28:26 | there everything else . We're gonna move it to the | |
| 28:29 | right side . So on the left side of the | |
| 28:32 | equation we're gonna have X squared , modest forex , | |
| 28:36 | leave a space negative eight wide . When we move | |
| 28:40 | it to the right side , it's gonna become positive | |
| 28:42 | eight y and negative four will change to positive . | |
| 28:46 | For now we're going to complete the square . So | |
| 28:50 | looking at this number , we're going to take half | |
| 28:52 | of it and then we're going to square . So | |
| 28:54 | I'm gonna add two squared to the left side . | |
| 29:00 | Two squared has a value of four since I've added | |
| 29:02 | four to the left side , I need to do | |
| 29:04 | the same to the right side such that the equation | |
| 29:07 | remains balanced . Now we can go ahead and complete | |
| 29:11 | the square . So it's gonna be x minus two | |
| 29:18 | squared on the right side , we can add four | |
| 29:23 | plus four , which will give us eight . So | |
| 29:25 | we have eight , Y plus eight . Next we | |
| 29:29 | can factor out an eight From eight , y plus | |
| 29:32 | eight . So this becomes y plus one . So | |
| 29:36 | now we have the equation of the problem in standard | |
| 29:42 | form . So I'm just going to erase what we | |
| 29:44 | have here and I'm just gonna rewrite a new equation | |
| 29:51 | here . So now that we have the standard form | |
| 29:58 | of the problem , we can go ahead and fine | |
| 30:02 | the rest of the answers we're looking for . So | |
| 30:04 | let's start with the coordinates of the vertex . Here | |
| 30:08 | we have negative two in front of X . We're | |
| 30:09 | going to change this to positive too . Here we | |
| 30:11 | have plus one . We're going to make this negative | |
| 30:14 | one . So that's the coordinates of the vertex . | |
| 30:17 | Hs two K is negative one . Now let's calculate | |
| 30:22 | P We're going to set eight equal to four p | |
| 30:27 | . So if eight is equal to four , p | |
| 30:29 | . eight divided by force to so we get that | |
| 30:32 | P Is equal to two . Now let's go ahead | |
| 30:39 | and sketch a graph . Most of the graph will | |
| 30:52 | be on the right side . So let's begin by | |
| 31:11 | plotting the vertex . So it's at 2 -1 , | |
| 31:15 | which is here . And then P is too . | |
| 31:20 | So what we need to do is determine what direction | |
| 31:23 | the probable is going to open . Is it going | |
| 31:25 | to open up down right or left ? What would | |
| 31:31 | you say ? Well , looking at this X is | |
| 31:35 | squared . Why is not ? So when you have | |
| 31:38 | a general graph , y equals X squared . That's | |
| 31:41 | a problem that opens up if he is positive which | |
| 31:44 | it is . So we know the graph is gonna | |
| 31:47 | go in this general direction . So now we know | |
| 31:50 | where the focus is going to be PS two . | |
| 31:53 | So going to units above the Vertex takes us to | |
| 31:56 | this point . So that's where the focus is . | |
| 32:00 | So we can say that the focus Has an x | |
| 32:04 | . value of two And the Y Value of one | |
| 32:09 | . Now let's confirm that with this equation . The | |
| 32:13 | equation for the focus when dealing with a vertical problem | |
| 32:15 | , that's one that opens up or down . It's | |
| 32:18 | gonna be a church comma K plus P . So | |
| 32:24 | we know HS two K is negative one plus PPS | |
| 32:29 | to -1 Plus two is 1 . So we've confirmed | |
| 32:35 | that this answer is correct . So that's the coordinates | |
| 32:39 | of the focus . Now let's find some other points | |
| 32:49 | . So to get the other points we need to | |
| 32:50 | travel to p units to the right . two P | |
| 32:55 | units to the left . So p is 2 . | |
| 32:58 | 2 . P is going to be four . So | |
| 33:00 | going four units to the right of the focus takes | |
| 33:03 | us to this point and four units to the left | |
| 33:07 | takes us to that point . So now we can | |
| 33:09 | sketch a rough graph of the hyperbole . It doesn't | |
| 33:14 | have to be perfect , but at least this will | |
| 33:17 | be a good approximation . So now let's find the | |
| 33:22 | equation of the metrics . So let's go pee units | |
| 33:25 | in the other direction . And so the direction , | |
| 33:30 | I mean the directorate's rather we'll be here . So | |
| 33:36 | this is at a y value of -1 . And | |
| 33:39 | here this is at a Y . Value of negative | |
| 33:41 | three . So we can say that the equation of | |
| 33:45 | the rhetoric six is why Is equal to -3 . | |
| 33:52 | Now let's confirm it with the equation for the rhetoric | |
| 33:56 | when dealing with a vertical problem . It's why is | |
| 33:59 | equal to k minus P . K . is negative | |
| 34:02 | one minus P . Which is positive too -1 -2 | |
| 34:07 | is -3 . So you want to be familiar with | |
| 34:10 | these equations . It's good to know how to use | |
| 34:13 | them . So that's why I want to go over | |
| 34:16 | confirming our answer with those equations . I want you | |
| 34:18 | to be familiar with those equations for the different types | |
| 34:21 | of problems that you may encounter . So that's basically | |
| 34:26 | it for this problem . We did everything that we | |
| 34:29 | needed to do . |
Summarizer
DESCRIPTION:
This video tutorial provides a basic introduction into parabolas and conic sections. It explains how to graph parabolas in standard form and how to graph parabolas with the focus and directrix. The vertex of the parabola can be identified by analyzing the equation in standard form.
OVERVIEW:
Finding The Focus and Directrix of a Parabola - Conic Sections is a free educational video by The Organic Chemistry Tutor.
This page not only allows students and teachers view Finding The Focus and Directrix of a Parabola - Conic Sections videos but also find engaging Sample Questions, Apps, Pins, Worksheets, Books related to the following topics.
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