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. 9x² + 16y² – 18x + 64y – 71 = 0

Accepted Answer

Hello, today we're gonna be graphing the equation of the ellipse and identifying its foci. So what we are given is 36 X squared plus four Y squared minus 100 and 44 X plus 56 Y plus 100 and 96 is equal to zero. Now, the first thing we want to do is to get this into an identifiable form of an equation of the ellipse. In order to do that, the first thing we can do is group up our X terms together, group up the Y terms together and move any constant to the right hand side of the equation doing this is going to give us 36 X squared minus X plus four Y squared plus 56 Y is equal to negative 196. Now, one thing to note is that we can factor out a 36 from the X group and factor out a four from the Y group. If we do this, we can rewrite the equation as 36 times the quantity X squared minus four X plus four times the quantity Y squared minus plus 14 Y is equal to negative 196. Now, what we want to go ahead and do is rewrite the X and Y groups as factor trinomial. In order to do that, we can go ahead and complete the square for the X group. And the Y group completing the square on the X group is going to use the second term negative four. And the third term that we're going to be adding is going to be negative four divided by two squared. And that is going to give us positive four and completing the square on the Y group is going to give us 14. So completing the square on the Y group is going to be 14 divided by two squared, which is going to give us positive 49. So what we're going to do is we're going to add these values to the X and the Y group and we can rewrite the equation as 36 times X squared minus four X plus four plus four times Y squared plus 14, Y plus 49 is equal to negative 196. Now, one thing to note is that we need to balance the equation since we added a four and a 49 to the left hand side of the equation, we must also add those same values to the right hand side. But we need to be careful with this process because keep in mind we factored out a 36 from the X group. And we factored out a four from the Y group. So we also need to include those factors with the respective values that we added. And that means that we're going to be adding 36 times four and four times 49 to the right hand side of the equation. So if we factor the X group, the X group is going to factor to give us X minus two squared and factoring the Y group is going to give us Y plus seven squared. And if we add and multiply all the values on the right hand side together, this is going to simplify to give us 144. So the equation of the ellipse is going to simplify to give us times X minus two squared plus four times Y plus seven squared is equal to 144. Now, the last thing we need to do is to set the equation equal to one. And in order to do that, we're gonna divide every term in this equation by 144. That is going to simplify the equation to give us X minus two squared over four plus Y plus seven squared, over 36 is equal to one. So this is going to be the equation of the ellipse that we're going to be graphing. Now, what we need to do is we need to come up with the shape of the ellipse. So just to recap the equation of the ellipse we came up with is X minus two squared over four plus Y plus seven squared, over 36 is equal to one. One thing to note is that the denominator underneath the Y term is greater than the denominator underneath the X term. And that means that the major axis of this S ellipse is going to be the Y axis. That means that the ellipse is going to have the major vertices with respect to the Y axis and minor vertices with respect to the X axis. We represent the distance of the minor vertices with the B variable and the distance of the major vertices with the X with a variable. And one thing to note is that we have quantities of X and Y in the numerators. And what this means is that this ellipse is not centered at the origin and the center is going to be defined as H comma K. Now that we know that the major axis is the Y axis and the center is not the origin. That means that the form of the equation of the ellipse is going to be 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 identify the A and B distances and identify the center of the ellipse. Now using the general form of the equation of the ellipse, we're going to set A squared equal to 36 and we're going to set B squared equal to four. If we take the square root of both of these equations, we get A is equal to plus or minus six and B is equal to plus or minus two. What this means is that the major vertices are located at distance of six above and below the center of the ellipse. And the minor vertices are located distance of two to the left and right of the center of the ellipse. Now let's go ahead and identify the center of the ellipse. If we take a look at the X group, we have X minus two squared. And this is in the general form of X minus H squared. This is where H is going to equal to two. And if we take a look at the Y term, we have Y plus seven squared, this is not in the general form of Y minus K squared, but we can rewrite this to give us Y minus negative seven squared. This is in the general form of Y minus K squared. And this is where K is equal to negative seven since we know that HS two and K is negative seven, that means the center of the ellipse is going to be located at two comma negative seven. So now that we have the center of the ellipse and we have the distances of the major and minor vertices. Let's go ahead and graph the lips. So we know that the ellipse is gonna be located or at least the center is gonna be located at two comma negative seven. Now, we know that the minor vertices are two distances to the left and right of the center of the ellipse. That means that the minor vertices are gonna be located at four comma negative seven and zero, comma negative seven. And the major vertices are gonna be located six units above and below the, the center of the ellipse. So we're going to be adding six to the Y value, which is going to give us a point of two comma negative one. And it's gonna be located six distance below the center which is going to give us two comma negative 13. So this is going to be the rough shape of our lips. Now, what we need to do is we need to identify the, now the foci are always located between the center and the major vertices and we represent the distance of the using ac variable. Now, for ellipses, we do have an equation that helps us identify the distance for the ellipse or for the size that is going to be C squared is equal to A squared minus B squared. And we already know what the values of A squared and B squared are going to be A squared is 36 and B squared is four. So that means that C squared is going to equal to minus four. If we simplify this, we get C squared is equal to 32. And if we take the square root and simplify, we get C is equal to four square root of two, so that means that the foci are gonna be located a distance of four square root of two above and below the center of the ellipse. And that means that we're going to be adding and subtracting four square root of two to the Y value of the center of the ellipse in order to get the locations of the foci. And that means that the are going to be located at two comma negative seven plus four square root of two and two comma negative seven minus four square root of two. So now that we have the values of the sign and we have the rough shape of the ellipse, we just need to identify the graph above that accurately represents this information that graph is going to be option a option A has the same equation of the ellipse that we came up with earlier. And it has the same values of the, the are two comma negative seven minus four square of two and two comma negative seven plus four square root of two. And with that being said, the answer to this problem is going to be a. 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. 9x² + 16y² – 18x + 64y – 71 = 0