SOLUTION: Find the vertices and the foci of the ellipse given by x^2+9y^2-24z+18y+9=0 (x squared plus 9y squared minus 24z plus 18y plus 9 equals 0)

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Question 249614: Find the vertices and the foci of the ellipse given by
x^2+9y^2-24z+18y+9=0
(x squared plus 9y squared minus 24z plus 18y plus 9 equals 0)
Answer by jsmallt9(3758)   (Show Source): You can put this solution on YOUR website!
I'm assuming that there is no "z". (If there is a "z", then this is a 3-dimensional graph, not an ellipse.)

We will start by transforming the equation into one of the following forms:
or
In this form we can get the center of the ellipse: (h, k). We can also get the values for a and b. And once we have the values for a and b we can find the value of c. And with the values for h, k, a, b and c we can find the vertices and the foci.

Since the desired form has both x and the y in a perfect square, we will complete the squares for both. To complete the squares we will gather the x terms and y terms together on one side of the equation and move the constant term to the other side. So I am going to change the subtractions to additions (so I can rearrange the terms) and subtract 9 from each side:

Next, I'll factor out the leading coefficient of the y terms, 9:

Next we want to find the number we need to add to the x terms and the number we need to add to the y terms to turn them into perfect squares:
(x^2 + (-24x) + ?) + 9(y^2 + 2y + ??)
The perfect square pattern is

We have the perfect square first terms, and . We have middle terms. We just need to figure out the correct third terms so each set of three terms fits the pattern above. With 1's in front of our first terms, the correct third term will be square of half the the middle coefficient. For the x's, half of the middle term is -12. The square of -12 is 144. For the y's, half of the middle coefficient is 1. The square of 1 is 1. So we want the left side of the equation to look like:

In order to make the left side of the equation look like this, we need to add the correct amount to both sides of the equation. But, and this is the tricky part, we will not be adding 144+1 to each side! The 1 is inside the parentheses which means anything we add inside the parentheses get multiplied by 9. So what we will be adding to both sides is 144+9*1 = 153. Now we have:

Now we rewrite the equation with perfect squares on the left and a simplified right side:

Since the desired forms have subtractions in the parentheses, I'll rewrite the left side using subtractions:

The last piece is to get the 1 on the right side. Dividing both sides by 144 does this:

The second fraction on the left will simplify:

We have the desired form. From this we can read:
h = 12
k = -1
which means a = 12
which means b = 4

Before we find c, let's review what a, b and c represent:
Next we want to find c. To find c we use an equation that describes the relationship between a, b and c. This equation we use is similar to the equation the describes the relationship between a, b and c in a hyperbola. These two equations are and . If you have trouble remembering which equation goes with which conic section, as I do, you can figure out which is correct logically. Thinking about what "a" and "c" represent and thinking about ellipses and where we find the foci and vertices, we can deduce that since the foci are inside the ellipse, a > c. With similar logic we can deduce that c > a for hyperbolas. The equation in which a > c is so this is the equation for ellipses (and is the equation for hyperbolas).

Using this equation and the values we have for and we get:

Solving for c we get:



The only information we still need is: Is this a horizontally-oriented ellipse (IOW is the major axis a horizontal line) or is it a vertically-oriented ellipse. This we get from the desired form. Since for ellipses (after all if b > a then b would be on the major axis), the is under the x term, this is a horizontally-oriented ellipse.

We now have all the information we need to answer the question.
Center.
For an equation in the desired form, the coordinates of the center of the ellipse is (h, k). So the coordinates of the center are:
(12, -1)
(Note: even though the problem does not ask for the center, we need it anyway. a, b and c are distances from the center.

Vertices on the major axis.
These are a distance of "a" from the center along the major axis. The major axis for this ellipse is horizontal so the vertices will be "a" units to the left and right of the center:
(12+12, -1) and (12-12, -1)
or
(24, -1) and (0, 1)

Vertices on the minor axis.
These are a distance of "b" from the center along the minor axis. The major axis for this ellipse is horizontal which makes the minor axis vertical. So these vertices will be "b" units to the up and down from the center:
(12, -1+4) and (12, -1-4)
or
(12, 3) and (12, -5)

Foci.
These are a distance of "c" from the center along the major axis. The major axis for this ellipse is horizontal so the vertices will be "c" units to the left and right of the center:
(, -1) and (, -1)
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