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. 36x^2 +9y^2 - 216x = 0

Accepted Answer

Hello, today we're gonna be graphing the equation of the ellipse and identifying its foci. So the equation of the lips given to us is X squared plus nine Y squared minus 18, Y minus 27 is equal to zero. The first thing we want to do is to go ahead and group up all of our X and Y terms together and move any constant to the right hand side of the equation doing this allows us to rewrite the equation as X squared plus the group nine Y squared minus 18 Y is equal to 27. Now notice that we can factor out a nine from the Y group doing this allows us to rewrite the equation as X squared plus nine times the quantity Y squared minus two Y is equal to 27. Now, what we want to do is complete the square on the Y group, that way we can rewrite it as a factor trinomial. Now, the equation for completing the square is given to us as C is equal to the middle term B divided by two squared. In this case, here B is going equal to negative two. So the third term C is going to be negative two divided by two squared. And that is going to simplify to give us one. So we can rewrite the equation as X squared plus nine times the quantity Y squared minus two Y plus one is equal to 27. Now, one thing to note is that we need to balance the equation. Since we added a one to the group on the left hand side, we must also add a one to the right hand side of the equation. But here's where the tricky part comes in. Keep in mind that we factored out a nine from the Y group. So we need to add one times that factor of nine to the right hand side. So on the right hand side, we're going to have 27 plus nine times one. So if we factor the left hand side and simplify the right hand side, the equation is going to simplify to give us X squared plus nine times the quantity Y minus one squared is equal to 36. And now what we wanna do is set the equation equal to one to represent this as an equation of ellipse. We're gonna do that by dividing everything by 36. And that is going to give us X squared over plus the quantity Y minus one squared over four is equal to one. This is going to be the equation of the ellipse that we want to graph. So let's go ahead and continue this work in the bottom. We currently have the equation X squared over plus Y minus one squared over four is equal to one. One thing to note is that this is the graph of an ellipse not centered at the origin because we have the quantity Y minus one squared. Let's go ahead and identify the major axis for this ellipse. The denominator underneath the X term is greater than the denominator underneath the Y term. So that means that the major axis for the ellipse is going to be the X axis. The ellipse is going to have the major vertices located with respect to the X axis and minor vertices with respect to the Y axis. And the center is going to be represented as H comma K. And since we know that this is in the equation of ellipse not located at the origin. The form of the equation of the ellipse with respect to the X axis is going to be X minus H squared over A squared plus Y minus K squared over B squared is equal to one. Now, this is where the A terms represent the values of the major X vertices. And the B terms represent the value of the minor vertices. So let's go ahead and first determine the distance of A from the origin or from the center of the ellipse, we're going to set A squared equal to the bigger denominator which is 36. And we're going to set B squared equal to the smaller denominator, which is four. If we take the square root of both sides of A, we get A is equal to plus or minus six. And taking the square root of the B equation is going to give us B is equal to plus or minus two. So that means that the major vertices are gonna be a distance of six away from the center and the minor vertices are going to be a distance of two away from the center. Now let's go ahead and determine the center of the ellipse. Well, if we take a look at the X term, we only have X squared, this is equivalent to X minus zero squared, which is in the standard form of X minus H squared. This is where H is going to equal to zero. And if we take a look at the Y term, we have Y minus one squared, which is in the standard form of Y minus K squared, this is where K is going to equal to one. And that means that the center of the ellipse is gonna be located at zero comma one. So let's go ahead and graph all of the given information. So we know that the center of the ellipse is going to be at zero comma one. And the major vertices are going to be a distance of six away from the center. So that means that the major vertices are gonna be located at six comma one and negative six comma one. And the distance of the minor vertices are gonna be two away from the center. So two units above the center of the ellipse is going to be at zero comma three and two away from the center below is going to be at zero comma negative one. So this is going to be the rough shape of the ellipse. Now, there's one more thing that we need to identify and that's going to be the of the ellipse. Now, the foci are located with respect to the major vertices. So they're gonna be some distance between the center and the major vertices. But how do we identify the ci well, we represent the foci with the C variable and the equation to sulfur the foci of an ellipse is going to be C squared is equal to A squared minus B squared. Now, we already have the values for A squared minus B squared. A squared is equal to 36 and B squared is equal to four. So that means that C squared is equal to 36 minus four, which is going to give us 32. And now we wanna sell for C if we take the square root of both sides, C is equal to the square root of 32. Now, what this means is that the foci are located some distance of square root of 32 away from the center of the ellipse. That means that we're going to be adding the distance of 32 and subtracting a distance of square root of 32 to the X value of the ellipse. And that means that the vertices, I'm sorry, the foci are gonna be located at square root 32 comma one and native square root 32 comma one. So now that we have the, and we have the rough shape of the ellipse, we just need to select the option that accurately represents all this information. And that option is going to be D. So that means that the answer to this problem is 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. 36x^2 +9y^2 - 216x = 0