Difficulties with ellipse: center, foci, major & minor axis?

Question

4x^2 + 9y^2 - 8x + 36y + 4 = 0
I know I have to move the 4 over
4x^2 + 9y^2 - 8x + 36y = -4
And then factor:
4(x^2 - 2x) + 9(y^2 + 4y) = -4
and find (-2/2)^2 & (4/2)^2 = 1 & 4 (add to left side and 4 & 36 to the right).

Where do I go from here and how to I find the center, foci, major & minor axis?

I really appreciate ANY help.
Thank you very much and Merry Christmas!

I'm sorry you had to do all that work! I didn't realize! Thank you so much, Philio!

Have a great Christmas! Take care :)

Answer 1

Well, I'm not sure how you've been taught, but here's how I'd do it. We want to get it into the form

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

Once in this form, the center, foci and axes are easy to find:

center = (h,k)

If a > b, major axis is x=h, minor is y=k, and foci are at (h+c, k) and (h-c, k)

If b > a, major axis is y=k, minor is x=h, and foci are at (h, k+c) and (h, k-c)

where c is found using the equation: c^2 = a^2 - b^2.

So to get the equation into the right form, do these steps:

Move the 4 to the right.

4x^2 + 9y^2 - 8x + 36y = -4

Factor.

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

Use completing the square to get

(x^2 - 2x) = (x - 1)^2 - 1
&
(y^2 + 4y) = (y + 2)^2 - 4

So substituting these into the original equation, we have

4[(x - 1)^2 - 1] + 9[(y + 2)^2 - 4] = -4

Distribute the 4 and 9:

4(x - 1)^2 - 4 + 9(y + 2)^2 - 36 = -4

Bring the -4 and -36 to the right:

4(x - 1)^2 + 9(y + 2)^2 = 36

Now divide everything by 36:

(1/9)(x - 1)^2 + (1/4)(y + 2)^2 = 1

So a^2 = 9, b^2 = 4, which means a = 3, b = 2.

The center is at (h,k) = (1,-2) and our major axis is x = 1, minor is y=-2 since a > b.

To find c, we have c^2 = 9 - 4 = 5, so c = sqrt(5).

So our foci are

(h+c, k) = (1+sqrt(5), -2)
&
(h-c, k) = (1-sqrt(5), -2)

Answer 2

4x^2 + 9y^2 - 8x + 36y + 4 = 0
4(x^2 - 2x + 1 - 1) + 9(y^2 + 4y + 4 - 4) + 4 = 0
4(x - 1)^2 - 4 + 9(y + 2)^2 - 36 + 4 = 0
4(x - 1)^2 + 9(y + 2)^2 = 36
Dividing by 36,
(1/9)(x - 1)^2 + (1/4)(y + 2)^2 = 1
center is (1, - 2)
a = 3, major axis is (- 2, - 2) to (4, - 2)
b = 2, minor axis is (1, - 4) to (1, 0)
c = √(9 - 4) = √5
foci are (1 - √5, - 2), (1 + √5, - 2)

Answer 3

4x² + 9y² - 8x + 36y + 4 = 0
(4x² - 8x) + (9y² + 36y) = -4
4(x² - 2x) + 9(y² + 4y) = -4
4(x² - 2x + 1) + 9(y² + 4y + 4) = -4 + 4 + 36
4(x - 1)² + 9(y + 2)² = 36

(x - 1)² .. (y + 2)²
-------- + ---------- = 1
... 9 .......... 4

so center is (1,-2),
major axis endpoints are ±√9 left and right of center at (-2, -2) and (4, -2)
minor axis endpoints are ±√4 up and down from center at (1, 0) and (1, -4).
foci are √5 (which is 9-4) left and right of center at (1-√5, -2) and (1+√5, -2)