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This video. Were gonna focus on conic sections circles ellipses hyperbolas and parabolas now now im gonna give you all the formulas that you need in order to the conic sections. And this video is just a simple basic introduction into the topic.
Im gonna have some other videos that explain how to graph each of these different types of conic sections and im gonna post the links to those videos in the description section below so feel free to take a look at that when you get a chance. The first thing that we need to talk about is a circle heres the general formula for the circle. Its x minus h.
Squared. Plus. Y minus k.
Squared is equal to r squared. So the first thing we need to do is you need to identify the center. Which is h.
Comma k. And then you need to calculate the radius. Now because graphing circles is very simple lets do an example on this one.
So. Lets say. We have the equation.
X. 3. Squared.
Plus. Y. Plus.
4. Squared. Lets say thats equal to 9.
So h. To find it simply change negative. 3 to positive 3 to calculate k change positive.
4 to negative. 4. So thats the center its 3 comma negative.
4. The radius is the square root of that number if r squared is 9 r. Is going to be the square root of 9 or 3.
Now once you know the center and the radius you can plot the circle. So lets begin by plot in the center. We need to travel 3 units to the right and then we need to go down 4 units.
So its here now the radius is 3 to get the endpoints of the circle. We need to go up 3 units three units to the left. Then we need to go down three units from the center and then we need to travel three units to the right so those are the endpoints of the circle and then we simply need to connect them through a circle.
Now my graph is not perfect. But those are the basic steps that you would take in order to plot a circle flatten. An ellipse is very similar to this now.
Lets talk about ellipses for the one on the left. The formula that corresponds to it is x minus h. Squared over a squared plus y.
Minus k. Squared over b squared is equal to one for the one on the right. Its x minus h.
Squared. Over b. Squared.
As opposed to a squared. So. As you can see a and b.
Have been reversed for these two types of ellipses. Now the center for both of them is still h. Comma k.
And very similar to the center of a circle.
But the difference between an ellipse is its an uneven circle. Now a is always bigger than b. Thats how you can tell the two equations apart.
So if this value is larger thats going to be your a squared. If the number under x minus. H.
Squared is larger. Thats your a squared. A a is always bigger than b for ellipses.
Now a is the distance between the center and the major vertices. So these are the major vertices highlighted in white and this here is the access. I mean.
Its the left of the major axis. The major axis is always equal to 2a b is the distance between a center and the minor vertices and the minor vertices are highlighted in blue here the left of the minor axis is 2b now for the other one the major axis is horizontal. Here the major axis is vertical on the left barna right the major axis is horizontal.
And its always going to be equal to 2ei so once again a is a distance between a center and the major vertices. So aim is always the long part in the ellipse and a major vertices are now along the x axis or parallel to it the minor vertices are along the y axis. If the center is centered at the origin.
So this is going to be b. So its important to calculate a and b. When graphing ellipses because they can help you to find the vertices and thus plot.
The ellipse now another thing we need to take into account is the value of c c. Allows us to determine the foci. The foci is always along the major axis.
So lets put f for that the distance between the center and the focus or any one of the focus is going to be c. So for the ellipse on our right the major axis is horizontal. So the foci will be along the major axis and its going to be c units from the center.
So to calculate c you need to use this formula. Its a squared minus b. Squared.
Its not the pythagorean theorem a squared plus b squared. But a squared minus b. Squared.
So remember a is bigger than b. And also a is bigger than c. If you get a c value.
Thats greater than a you know you made a mistake. C. Has to be less than a now in order to determine the coordinates of the foci heres what we need to do or his work can help you to get it for whenever.
The axis is the major axis is vertical. Its going to be h. Comma k.
Plus. Or minus c. Now lets understand why remember see is along the vertical axis.
Its parallel to the y axis. So thus you want to add. And subtract c.
To the y coordinate of the center of the ellipse now for the ellipse on the right the major axis is horizontal or parallel to the x axis. So were going to add plus or minus c. To the x coordinate of the center.
So thats how you can determine the coordinates of the foci for these two different types of ellipse now lets talk about fine in the area of an ellipse. But first lets find the area of a circle. So this is the sun of the circle.
And this is the radius and this is also the radius. The area of a circle is pi r. Squared you could think of it.
As pi time is the first r. Times. The second r.
Now lets compare this to an ellipse.
Okay. Thats a terrible looking ellipse. Lets do that again all right well work with that so lets say this is the center.
This is going to be a and this is going to be be the major axis is horizontal. So the equation is going to be very similar to this one. But instead of pi times r.
Times r. R. Is not the same here.
Its different. Its gonna be pi times. A times b.
So thats how you could determine the area of an ellipse is very similar to the area of a circle. But because the circle is uneven you know its not going to be r times r. Its gonna be a times b the next type of graph we need to talk about is the hyperbola for the one on the left.
It corresponds to this formula x minus h squared over a squared minus y minus k squared over b squared is equal to 1 notice the difference between this and the ellipse with the ellipse. Theres a plus sign. But with the hyperbola.
Theres a minus sign now for the hyperbola on the right. The equation is y. Minus k.
Squared. Over a squared minus x. Minus h.
Squared over b squared is equal to 1. So. If x.
Squared. Is positive is going to open to the left and to the right. If a y squared is positive.
Its gonna open up and down. So. Thats how you determine the direction in which the hyperbola opens.
A is the distance between the center and the vertex now theres two vertex. So is the distance between the center and the vertices. So here are the vertices of the hyperbola to find the coordinates of the vertices you can use this its gonna be h.
Plus or minus a comma k. Because a is horizontal for the hyperbola on the left on the right. A is vertical.
A is the distance between the center and the vertices so in this case. The coordinates of the vertices will be h. Comma k.
Plus. Or minus a now the full side you could find them where the hyperbola opens up towards it always opens towards the foci and see not the center c. But the value c is the difference between the center and the foci so to find the coordinates of the foci.
Its going to be h. Plus. Or minus c.
Comma k. Now for that per beulah on the right c is vertical. So were gonna add and subtract c from k.
And this should be an f. Not a v. So thats how you could find the coordinates of the foci in wendell.
If i hyperbola now the last thing we need to talk about is finding the equation of the asymptotes. So let me clear away a few things so when flatten a hyperbola you would find the center travel. A units to the right to get the vertex on the right a units are left.
And then you would also travel b units up and down. This is just a rough sketch by the way and then once you have a and b. What you want to do is you want to plot a rectangular dotted box.
And then what youre supposed to do is you apply.
The asymptotes. Which goes from one diagonal of the box through the center to the other corner of the box. So you draw these diagonals that go from one corner of the box to the other corner to the center and then to find the equation of those two asymptotes its gonna be y minus k.
Basically what you see here equals now the slope starting from the center. The slope is the rise over the run. So we rose b units and travel or ran a units to the right.
So the slope is going to be plus or minus. And b. Over a and then what we see here x minus h.
Now. The reason. We have plus or minus.
Is the first diagonal has a positive slope. So thats going to be the equation that corresponds to the positive b over a the second diagonal has a negative slope. So thats the second equation that corresponds to negative b over a thats what we have two because there are two asymptotes one with a positive slope and the other with a negative slope.
Now for the hyperbola on the right the equation of the asymptotes will be y minus k. Equals. Plus or minus.
Notice the slope is different this time the rise is a the run is going to be b. So instead of plus. A minus b over a its now plus or minus a over b.
And then x minus h. So thats the difference between the asymptotes of these two types of hyperbolas. The difference is in the slope in one case.
Its b over a any other case. Its a would be if he ever forget. Its always gonna be the rise over the run now we talked about fine in the coordinates of the foci by calculating the value of c or use in c define.
It now we need to talk about how to find c with the ellipse. We said that a was larger than c. With the hyperbola c.
Is going to be larger than a so c. Squared is gonna be not a squared minus b. Squared.
But a squared plus b. Squared so for the hyperbola. This formula follows.
The pythagorean theorem. So thats how you could find c in order to find the coordinates of the foci. So thats basically it for the hyperbola.
This is all you need to know in order to graph it and answer questions associated with it and once again for those of you who want practice problems check out the links in the description section below. If you want to see some examples in terms of how to graph a hyperbola and other problems associated with our purpose. Now lets move on to parabolas so lets start with this equation.
Y minus k. Squared is equal to 4 p times x minus h. Now we can get two graphs associated with this equation for the first one the parabola can open towards the right you can think of the parabola.
As just one part of the hyperbola. It can also open towards the left the direction in which it opens depends on the p value when p is positive its gonna open towards the right if p is negative. Were gonna its gonna open towards the left so it really depends on what you see here if you were to see like for example negative.
8 p. Would be negative if you saw a positive 6 p is positive now for the other equation. Its gonna be x minus h.
Squared is equal to 4p times. Y minus k. And for this equation it can either open upward or it can open downward if it opens upward p is positive if it opens downward p is negative now you might be wondering what can help me to remember these things well look at the formula here we have x x.
Is not squared. But y is squared. So i think of this equation x.
Is equal to y squared. For this formula. Its basically a parabola that opens left and right for the other one x is squared.
Is not so we get this equation. Y. Is equal to x squared.
Youve seen. This one either in the form of a quadratic equation or parabola. If we have y equals positive x squared.
We know we have a parabola that opens upward. If its y equals the negative x squared. It opens downward.
So that can help you to remember the direction. So when you have x equals. Y.
Squared. It opens either to the left or to the right. But if you have y equals.
X. Squared. It opens up or down.
Now we need to talk about how to find the equation of the directrix and the focus. So the directrix is always away from wherever. The parabola opens up and the parabola it always opens towards the focus.
The directrix is p units away from the vertex and the focus is also p units away from the vertex. So one of the first things you need to do is find the center. Which is h.
Comma k. And then calculate the value of p. Because once you have p you could find the equation of the directrix and the coordinates of the focus.
So for the graph at the top left. The coordinates of the focus is going to be the center h. And then since we travel p.
Even to the right. Its gonna be h plus p. And then comma k for the next graph on the right the coordinates of the focus is going to be h minus p.
Because we went p units to the left from h and then comma k. So thats the coordinates of the focus for the graph on the right now for the equation of the directrix since we have a vertical line. Its going to be x equals.
So its x equals h. Minus p. Here.
We have the other equation of the directrix and its going to be x is equal to h plus p. Since. We travel p units to the right from the center.
So these are the equations of the directrix for the first two graphs now lets move on to the bottom ones. So the equation of the directrix is going to be different the directrix is now horizontal. So instead of x equals is going to be y equals for this one.
Its going to be y equals. K. Minus p.
And for the other one its y equals. K. Plus p.
So remember the directrix is always away from the direction in which the parabola opens towards the focus is always where its in the direction where the parabola opens. Towards so the focus for the one at the bottom left is going to be h. Comma.
K. Plus p. And for the other one its h.
Comma k. Minus p. So now you know how to find the equation of the directrix and the coordinates are the focus when dealing with parabolas so thats basically it for this video for those of you who want practice problems again check out the links in the description section below.
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