Find the center of an ellipse with the equation
.
You first have to put it in standard form, which is either
(x - h)² (y - k)²
---------- + ---------- = 1
a² b²
if it turns out that the larger number is under
the term on the left which contains x, and the
graph will be an ellipse with a horizontal major
axis and a vertical minor axis, where
1. The center is (h, k)
2. The length of the semi-major axis is " a ".
3. The ends of the major axis are the points (h-a,k) and (h+a,k)
4. The ends of the minor axis are the points (h,k-b) and (h,k+b)
5. Foci are (h-c,k), (h+c,k) where c is calculated from c² = a²-b²
or
(x - h)² (y - k)²
---------- + ---------- = 1
b² a²
if it turns out that the larger number is under
the term on the left which contains y, and the
graph will be an ellipse with a horizontal major
axis and a vertical minor axis, where and the
graph will be an ellipse with a vertical major
axis and a horizontal minor axis
where
1. The center is (h, k)
2. The length of the semi-major axis is " a ".
3. The ends of the major axis are the points (h,k-a) and (h,k+a)
4. The ends of the minor axis are the points (h-b,k) and (h+b,k)
5. Foci are (h,k-a), (h,k+a) where c is calculated from c² = a²-b²
We can't tell which it is unti we get it into standard form:
9x² + 16y² - 18x + 64y = 71
Rearrange getting x² term then the x term then y²and y terms
That is, swap the second and third terms:
9x² - 18x + 16y² + 64y = 71
Factor just the 9 out of the first two terms
on the left and also factor just the 16 out of
the last two terms on the left:
9(x² - 2x) + 16(y² + 4y) = 71
Complete the square in the first parentheses:
The coefficient of x is -2
Multiply it by 
, getting -1
Square -1, getting +1.
Add +1 inside the first parentheses.
But the parentheses has a 9 in front.
So this amounts to adding 9·1 or 9 to
the left side, so we must add +9 to
the right side:
9(x² - 2x + 1) + 16(y² + 4y) = 71 + 9
Complete the square in the second parentheses:
The coefficient of y is 4
Multiply it by , getting 2
Square 2, getting +4.
Add +4 inside the second parentheses.
But the parentheses has a 16 in front.
So this amounts to adding 16·4 or 64 to
the left side, so we must add +64 to
the right side:
9(x² - 2x + 1) + 16(y² + 4y + 4) = 71 + 9 + 64
Factor both parentheses on the left and combine the
numbers on the right
9(x - 1)(x - 1) + 16(y + 2)(y + 2) = 144
9(x - 1)² + 16(y + 2)² = 144
To get a 1 on the right side, divide every term
by 144
9(x - 1)² 16(y + 2)² 144
--------- + ---------- = -----
144 144 144
Cancel
1 1 1
9(x - 1)² 16(y + 2)² 144
--------- + ---------- = -----
144 144 144
16 9 1
(x - 1)² (y + 2)²
---------- + ---------- = 1
16 9
Since 16 is greater than 9, and since a² is always the
larger of these in an ellipse, then a² is under the
term which contains x, we compare that to:
(x - h)² (y - k)²
---------- + ---------- = 1
a² b²
so it is an ellipse with a horizontal major axis and
a vertical minor axis. Comparing letters in this with
the numbers in the equation we got, we see that:
h = 1, k = -2, a² = 16, b² = 9 so a = 4 and b = 3
This means
1. The center is (h, k) = (1, -2)
2. The length of the semi-major axis is a = 4
3. The ends of the major axis (vertices) are the points (h-a,k), (h+a,k)
or (1-4,-2) and (1+4,-2) or (-3,-2) and (5,-2)
4. The ends of the minor axis are the points (h,k-b), (h,k+b)
or (1,-2-3), (1,-2+3) or (1,-5) and (1,1)
5. Foci are (h-c,k), (h+c,k) where c is gotten from c² = a²-b²
We calculate c:
c² = a² - b²
c² = 4² - 3²
c² = 16 - 9
c² = 7
c = Ö7 _ _
So foci are (h-c,k) and (h+c,k), or (1-Ö7,-2) and (1+Ö7,-2),
or about (-1.6,-2) and (3.6,-2)
We draw the major and minor axes and plot the foci:
Then we sketch in the ellipse:
The answers to your problem are:
center = (1, -2)
vertices = ends of major axis = (-3,-2) and (5,-2)
foci = (1-Ö7,-2) and (1+Ö7,-2),
Edwin
