In Exercises 51–60, convert each equation to standard form by com...

Question

In Exercises 51–60, convert each equation to standard form by completing the square on x and y. Then graph the ellipse and give the location of its foci. 4x² + y²+ 16x - 6y - 39 = 0

Accepted Answer

Hello, today we're gonna be graphing the equation of the ellipse and identifying the location of its foci. So what we are given is four X squared plus Y squared plus 64 X minus two Y plus is equal to zero. Now, currently are the equation of the ellipse is not in an identifiable form of an equation of an ellipse. But what we can go ahead and do in order to get that form is to group up our X and Y terms together and move any constant to the right of the equation doing so is going to give us four X squared plus 64 X plus Y squared minus two Y is equal to negative 221. Now, one thing to note is that we can go ahead and factor it out a factor of four from the X group. And if we do that, we can rewrite the equation as four times the quantity X squared plus 16 X plus Y squared minus two Y is equal to negative 221. Now, the next step into getting the equation of the ellipse is to go ahead and transform our X and Y groups into factor trinomial. In order to do that, we're gonna go ahead and complete the square on both of these quantities. So for the X group, we're going to be completing the square using the second term 16 X, that is going to give us 16 divided by two squared, which is going to give us 64. And if we complete the term of complete the square of the Y terms by using negative two, what we get is negative two divided by two squared, which is going to give us one. Now, what we can go ahead and do is add those quantities into the X and Y groups. And we can rewrite the equation as four times X squared plus 16 X plus plus Y squared minus two, Y plus one is equal to negative 221. Now, one thing to note is that we need to balance the equations since we added 64 and one to the left hand side of the equation, we must also add those same quantities to the right hand side of the equation. However, we want to be careful because for the X group, we factored out a four originally. So we also need to include that factor of four when adding the values to the right hand side of the equation. So on the right hand side, we're going to be adding four times plus one. And the reason why we're only adding one at the end is because we added a one on the left hand side, but we didn't factor out anything from the Y group originally. So let's go ahead and factor the X and Y groups. If we factor the X group, the group is going to factor to give us X plus eight squared and the Y group is going to factor to give us Y minus one squared. And if we multiply and add all the values together on the right hand side, this is all going to simplify to give us 36. So the equation is simplified to give us four times of the quantity X plus eight squared plus Y minus one squared equal to 36. Now we want to go ahead and set this equation equal to 0 to 1. And in order to do that, we can divide everything by 36. If we divide every turn by 36 the equation of the ellipse is going to simplify to give us X plus eight squared over eight. I'm sorry, over nine plus Y minus one squared, over 36 is equal to one. And this is going to be the equation of the ellipse that we're going to graph. So now let's go ahead and graph the ellipse down below. So again, the equation we came up with is X plus eight squared over nine plus Y minus one. Squared over 36 is equal to one. Now, there's a couple of things that we can identify from just looking at the equation of the ellipse. Notice that the denominator below the Y term is greater than the denominator below the X term. That means that the major axis of the ellipse is going to be the Y axis that tells us that the major vertices are gonna be located with respect to the Y axis. And the minor vertices are gonna be located with respect to the X axis. We represent the values of the minor vertices using A B variable. And we represent the values of the major vertices using an A variable. Furthermore, since we have X plus eight squared and Y minus one squared, this tells us that the equation of the ellipse is not centered at the origin and the center is going to be defined as H comma K. Now, since we know that the ellipse is not centered at the origin and the major axis is the Y axis, that means the form of the equation of the ellipse is in the form of X minus H squared over B squared plus Y minus K squared over A squared is equal to one. So let's go ahead and find your A and B variables and identify the center. So using the form of the equation of the ellipse, we're going to set A squared is equal to and we're going to set B squared is equal to nine. If we take the square root of the A equation, we get A is equal to plus or minus six. And if we take the square root of the B equation, we get B is equal to plus or minus three. What this means is that the major vertices are going to be a distance of six away from the center of the ellipse. And the minor vertices are going to be a distance of three away from the center of the ellipse. Now, we need to identify the center of the ellipse. So let's take a look at our X term, we have X plus eight squared. Currently, this is not in the standard form of X minus H squared, but we can rewrite it in a standard form by rewriting the statement as X minus negative eight squared. Now this is in the form of X minus H squared and this is where H is equal to negative eight. And for the Y group, we have Y minus one squared, this is in the standard form of Y minus K squared and this is where K is equal to one. So that means that the center of our ellipse is going to be located at negative eight comma one. So now that we have the center of the ellipse and we have the locations of the major and minor vertices. Let's go ahead and graph this ellipse. So we know that the ellipse is going to be centered at negative eight comma one. And that means that this is going to be centered in the second quadrant roughly summer here. Now, we know that the minor vertices are located a distance of three to the left and right of the center of the ellipse. And what that means is that the minor vertices are gonna be located at negative five comma one and negative 11 comma one. And we know that the major vertices are located a distance of six above and below the center of the ellipse. That means that the locations of the ellipse are going to be at negative eight comma negative five and negative eight comma positive seven. So now that we have the shape of the ellipse, we just need to identify the sign. Now, now we know that the foci are located between the center of the ellipse and the major vertices. But how do we find the distance for the sign? Well, we represent using AC variable and we do have an equation that helps us find this value that's going to be C squared is equal to A squared minus B squared. And here we already know the values of A squared and B squared A squared is 36 and B squared is nine. So if we substitute this in for the C equation, we get C squared is equal to 36 minus nine. Now 36 minus nine is going to give us 27. And if we take the square root of both sides, we get C is equal to three times the square root of three. So that's going to be the distance that the foci are away from the center of the ellipse. And since the foci are located above and below the center of the ellipse, that means we're going to be adding and tracking the value of C to the Y value of the center of the ellipse. That means that the foci are going to be located at negative eight comma one plus three square root of three and negative eight comma one minus three square root of three. So that's gonna be the location of our. So now that we have the shape of the graph and we have the locations of the site, we just need to select the graph above that accurately represents this information. And that option is going to be D D contains the graph that we identified earlier and it contains the locations of the foci which are negative eight comma one minus three square root of three and negative eight comma one plus three square root of three. So with that being said, the answer to this problem is going to be D. So I hope this video helps you in understanding how to approach this problem. And I'll go ahead and see you all in the next video.

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In Exercises 51–60, convert each equation to standard form by completing the square on x and y. Then graph the ellipse and give the location of its foci. 4x² + y²+ 16x - 6y - 39 = 0