ellipse help for algebra?

Question

could you help me put this in the ellipse standard form, the major axis endpoints, minor axis endpoints, and the foci points of it ?

please

x^2+64y^2+8x-48 = 0

please please help

so, i think i might have gotten the the answer, so could you please tell me if it's right?

standard form i got:

(x+4)^2 / 64 + y^2 = 1

Major axis endpoints i got: (-4,0) & (-12, 0)
Minor axis endpoints i got: (-4,1) & (-4, -1)

and for the foci points: (-11.94,0) & (3.94,0)

is this right?

Answer 1

x² + 8x + 64y² = 48

Complete the square for x by adding (8/2)² = 16

x² + 8x + 16 + 64y² = 48 + 16 = 64
(x + 4)² + 64y² = 64

Divide both sides by 64 = 8² to get in standard form:

(x + 4)²/8² + (y+0)²/1² = 1

The center is (-4, 0).

The endpoints of the major axis would be (-4 + 8, 0) = (4, 0), and (-4 -8, 0) = (-12, 0).

The endpoints of the minor axis would be (-4, 0+1) = (-4, 1) and (-4, 0-1) = (-4, -1).

The focal length is √(a² - b²) = √(64 - 1) = √63 = 3√7

So the foci would be (-4 + 3√7, 0) and (-4 - 3√7, 0).

If I didn't mess up.

Answer 2

Complete the square first:

(x^2 + 8x + 16) - 16 + 64y^2 - 48 = 0

(x + 4)^2 + 64y^2 = 64

Now divide through by 64 to make the right side equal 1:

(x + 4)^2/64 + y^2 = 1

Thus, the center of the ellipse is (-4, 0).

The major axis has length sqrt(64) = 8, so the major axis endpoints are (-12, 0) and (4, 0).

The minor axis has length sqrt(1) = 1, so the minor axis endpoints are (-4, -1) and (-4, 1).

The distance c that the foci lie from the center is given by c^2 = 64 - 1, or c = sqrt(63). The foci lie on the major axis, so they are
(-4 - sqrt(63), 0) and (-4 + sqrt(63), 0).

Summary:
Standard form: (x + 4)^2/64 + y^2 = 1
Major axis endpoints: (-12, 0) and (4, 0)
Minor axis endpoints: (-4, -1) and (-4, 1)
Foci: (-4 - sqrt(63), 0) and (-4 + sqrt(63), 0).