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. 9x^2 +25y² - 36x + 50y – 164 = 0

Accepted Answer

Hello, today we're gonna be graphing the equation of the ellipse and identifying its. So what we are given is 49 X squared plus 16, Y squared minus 392 X plus Y plus 64 is equal to zero. Now, the first thing we want to do is to rewrite this into an identifiable form of an ellipse. In order to do that, we can group up the X terms together, group up the Y terms together and move any constant to the right of the equation doing this allows us to rewrite the equation as 49 X squared minus X plus 16 Y squared plus 64 Y is equal to negative 64. Now, one thing to note is that we can factor out 49 from the X group and factor out 16 from the Y group. If we do this, we can rewrite the equation as 49 times the quantity X squared minus eight X plus 16 times Y squared plus four, Y is equal to negative 64. Now, what we wanna do is we want to turn the X and Y groups into a factor trinomial. And we can do this by completing the square by completing the square, we can add a third term which we call ac term to each of the factor binomials. We'll take the second term, which in this case is negative eight. For the X group, we're gonna divide it by two and then square it negative eight divided by two squared is going to give us 16. And for the Y group, if we repeat this process, we're going to take positive four divided by two and square, that value and four divided by two squared is going to give us positive four. So if we add these values to the X and Y groups, we can rewrite the equation as 49 times X squared minus eight X plus 16 plus 16 times Y squared plus four, Y plus four is equal to negative 64. Now, one thing to note is that we need to keep the equation balanced. Since we added 16 and four to the left hand side of the equation, we must also add those same values to the right hand side of the equation. But we need to be careful because keep in mind we factored out a 49 from the X group and factored out a 16 from the Y group. So we have to include those factors when we're adding the extra values to the right hand side of the equation. And what that means is that we're going to be adding 49 times 16 plus 16 times four to the right hand side of the equation. Now, if we factor the X and Y groups and add all the values together, on the right hand side, we can rewrite the equation as 49 times the quantity X minus four squared plus 16 times the quantity Y plus two squared is equal to 784. Now, the last thing we wanna do is we want to set the equation equal to one. In order to do that, we're going to divide everything by 784. And that's going to allow us to simplify the equation as X minus four squared over plus Y plus two squared over 49 is equal to one. This is going to be the equation of the ellipse that we're looking for. Now, there's a couple things to note about this equation of the ellipse. First, the denominator underneath the Y group is greater than the denominator underneath the X group. And what that means is that the major axis of this ellipse is going to be the Y axis. That means that the major vertices are going to be with respect to the Y axis and the minor vertices are going to be respect to the X axis. We're going to represent the distance of the major vertices using an A variable and represent the distance of the minor vertices using a B variable next, because we have the quantities X minus four squared and Y plus two squared that tells us that the origin or the center of this ellipse is not located at the origin and the center of the ellipse is going to be defined as H comma K. And because we know that this is not located at the origin and it has a major axis of the Y axis. The form 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 using the general form of the equation of the ellipse, we can identify our A and B distances and find the center of the ellipse. So from the general form, we can set A squared is equal to 49 and we can set B squared is equal to 16. If we solve for A squared and B squared, we get A is equal to plus or minus seven and B is equal to plus or minus four. What this means is that the locations of the major vertices are gonna be seven units above and below the center of the ellipse. And the distance of the minor vertices are gonna be a distance of four to the right and left of the center of the ellipse. Now we need to identify the center of the ellipse. If we take a look at our X group in the equation of the ellipse, we have X minus four squared. What this is is in the standard form of X minus H squared. And this is where H is going to equal to four. And if we take a look at the Y group, we have Y plus two squared. This is not in the standard form of Y minus K squared, but we can rewrite it as Y minus negative two squared. Now, this is in the standard form of Y minus K squared. And this is where K is equal to negative two. Now that we have our H value and K value, this tells us that the center of the ellipse is going to be located at four comma negative two. So now that we have the center of the ellipse and the locations or the distances of the vertices, we now need to identify where the is going to be. Well, the foci is always located between the center of the ellipse and the major, the major vertices. And we represent that distance with AC variable. Well, for an ellipse, we have an equation to help us find the distance of the that is going to be C squared is equal to A squared minus B squared. In this case here, we already know what our A squared and B score values are going to be A squared is equal to 49 and B squared is equal to 16. So that means C squared. Is going to equal to 49 minus 16, 49 minus 16 is going to give us the value of 33. So we have C squared is equal to 33. And if we take the square root of both sides of the equation, we get C is equal to the square root of 33. So now that we know that the distance of the foci is the square root of and the foci are located between the center and major vertices. We're going to be adding and subtracting square root of 33 to the Y value of the center of the ellipse. And that means that the foci is gonna be located at four comma negative two plus the square root of 33 and four comma negative two minus the square root of 33. So now that we have the information to graph the ellipse and we have the locations of the sign, we just need to select the graph that accurately represents the information we came up with. And that graph is going to be option C option C has the equation of the ellipse that we calculated and it has the values of the side that we calculated. And with that being said, the answer to this problem is going to be C 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

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. 9x^2 +25y² - 36x + 50y – 164 = 0

Key Concepts

Completing the Square

Standard Form of an Ellipse

Foci of an Ellipse

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