I need help with this algebra 2 ellipse problem.?

Question

It says:
Find the center, foci, length of major axis and length of minor axis.
x²+9y²-4x+54y+49=0

How do put it in the ellipse form? Do I have to complete the square?

Answer 1

Yup, you gotta complete both sqaures, you have:

x^2 - 4x
y^2 + 54y

Completing these squares:

x^2 - 4x + 2^2 = (x - 2)^2
9y^2 + 54y + 9^2 = (3y + 9)^2 = 9(y + 3)^2

So:

x^2 - 4x = (x - 2)^2 - 4
9y^2 + 54y = 9(y + 27)^2 - 81

Combining:

x²+9y²-4x+54y+49= (x - 2)^2 - 4 + 9(y + 27)^2 - 81 + 49 = 0
(x - 2)^2 + 9(y + 27)^2 = 36
((x - 2)/6)^2 + ((y + 27)/2)^2 = 36

Answer 2

The easiest way is definitely to get this into ellipse form!

(x-h)^2/a^2 + (y-k)^2/b^2 = 1

x^2-4x+something is (x-h)^2 = x^2 - 2xh + h^2
so h needs to be 2

(y-k)^2 = y^2 + 54/9y + something, and since 54/9 = 6, k is 3

So now, (x-2)^2 + 9*(y+3)^2 = something, which can be found by looking at the constant terms:
+4 in the first term, and 9*9 in the second term, for a total of 85.
The difference between 95 and 49 is 36, so the equation becomes:
(x-2)^2 + 9*(y+3)^2 = 36
or, dividing both sides by 36,
(x-2)^2/6^2 + (y+3)^2/2^2 = 1

From this, we can note that the CENTER is at (2,-3)
This is a horizontally oriented ellipse, so the foci are on the same horizontal line as the center. The distance from x=2 to the foci is c = sqrt(a^2-b^2), so the foci are at
(2+sqrt(6^2-2^2), -3), and (2-sqrt(32), -3)
= (2+4sqrt(2), -3) and (2-4sqrt(2), -3)

The length of major axis is 6 and the minor axis, 2

Answer 3

You need to complete the squares.

x²+9y²-4x+54y+49=0
(x² - 4x + 4) + 9(y² + 6y + 9) + 49 - 4 - 81 = 0
(x - 2)² + 9(y + 3)² - 36 = 0
(x - 2)² + 9(y + 3)² = 36
(x - 2)²/36 + (y + 3)²/4 = 1

Semi-major axis = a = √36 = 6
Semi-minor axis = b = √4 = 2
The major and minor axes are twice the above (2a and 2b).

The center (h,k) = (2,-3)

c² = a² - b² = 36 - 4 = 32
c = √32 = 4√2

So the foci are at (2+c,-3) and (2-c,-3)
Foci = (2 + 4√2,-3) and (2 - 4√2,-3)

Answer 4

Yeah, you'd have to complete the square on both variables to put it in the standard ellipse form. Then when you have it in the standard ellipse form you can read the center, foci, etc, from that.

To complete the square, do this:
x^2 - 4x + ..... + 9y^2 + 54y + ..... = -49
x^2 - 4x + 4 + 9(y^2 + 6y + 9) = -49 + 4 + 81 = 36
(x-2)^2 + 9(y+3)^2 = 36
(x-2)^2/36 + (y+3)^2/4 = 1

Answer 5

x² + 9y² - 4x + 54y + 49 = 0
x² - 4x + 4 + 9y² + 54y +81 + 49 - 4 - 81 = 0
(x-2)^2 + (3y + 9)^2 = 36
(1/9)(x-2)^2 + (1/1)(y + 3)^2 = 4

vertices at (2, -1), (2, 5), (-4, -3), (8, -3)
Center = (2, -3)
a = 3
b = 1
c = 2√2
foci at (2 + 2√2, -3), (2 - 2√2, -3)
major axis = 6
minor axis = 2

Answer 6

(x-2)^2-4+9(y+3)^2-81+49=0
(x-2)^2/36+(y+3)^2/4=1

centre=(2,-3)
4^2=6^2(1-e^2)
1-4/36=e^2
32/36=e^2
e=2rt2/3
foci(+/-ae,0)
vertices(8,-3),(-4,-3)
and (2,-1),(2,-5)