For the ellipse with equation 5x^2+64y^2+30x+128y-211=0, find the cooridinates of the center, foci, and vertices. Then, graph the equation.

my answer is:
coordinates of center: (-3,1)
foci:(-11,-1) and (4.7,-1)
vertices: (-11,-1) (5,-1)
(-3,-1+-square root of 5)
not sure how to graph, i know the center is (-3,1)

5(x^2 + 6x + 9) + 64(y^2 + 2x + 1) -45 -64 -211 = 0

5(x+3)^2 + 64 (y+1)^2 = 320
(x+3)^2/64 + (y+1)^2/5 = 0

Shouldn't the center be at (-3, -1) ?
The semimajor axis length (along the x direction) is a = sqrt64 = 8 and the semiminor axis length is b = sqrt5. The foci are at y = -1, and x = -3 +/- c, where
c^2 = a^2 - b^2 = 64 - 5 = 59

Add and subtract the semimajor axis lengths from the center coordinate to get the vertex locations.

I can't help you with the graphing part. You will need to locate the center on a graph, plot the vertex locations, and compute the coordinates of some intermediate points along the curve.

It is possible that I have made some math errors myself, so check my work.

4x^2-9x+32x-144y-548=0

To find the coordinates of the center, foci, and vertices of the given ellipse equation, you need to rewrite the equation in a standard form. The standard form for an ellipse is:

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

where (h, k) represents the coordinates of the center, and "a" and "b" are the lengths of the semi-major and semi-minor axes, respectively.

1. First, let's rearrange the given equation and complete the square for both the x and y terms:
5x^2 + 64y^2 + 30x + 128y - 211 = 0.

Rearranging the terms, we get:
5x^2 + 30x + 64y^2 + 128y = 211.

2. Divide the entire equation by 211 to simplify it:
(x^2 + 6x) / 211 + (y^2 + 2y) / (211/64) = 1.

Now, we can see that the coefficients of x^2 and y^2 terms are both 1, which means the equation is already in the standard form.

3. Complete the square for the x terms:
(x^2 + 6x + 9) / 211 + (y^2 + 2y) / (211/64) = 1 + 9/211.

(x + 3)^2 / 211 + (y^2 + 2y) / (211/64) = (220/211).

4. Complete the square for the y terms:
(x + 3)^2 / 211 + (y^2 + 2y + 1) / (211/64) = (220/211) + 1/(211/64).

(x + 3)^2 / 211 + (y + 1)^2 / (211/64) = (220/211) + (64/211).

Rewriting to a common denominator and simplifying:
(x + 3)^2 / 211 + (y + 1)^2 / (211/64) = (220 + 64) / 211.

(x + 3)^2 / 211 + (y + 1)^2 / (211/64) = 284 / 211.

Now we have the standard form equation of the ellipse.

5. From the equation, we can determine that:
- The center of the ellipse is at (-3, -1), corresponding to the values of h and k.
- The semi-major axis "a" can be found by taking the square root of the fraction in front of (x + 3)^2, i.e., sqrt(211).
- The semi-minor axis "b" can be found by taking the square root of the fraction in front of (y + 1)^2, i.e., sqrt(211/64).

6. The foci can be obtained by using the formula:

c = sqrt(a^2 - b^2),

where "c" represents the distance from the center to the foci.

By substituting the values of "a" and "b" into the formula, we get:
c = sqrt(211 - 211/64) = sqrt(13584/64 - 211/64) = sqrt(13373/64) ≈ 9.72.

Thus, the foci of the ellipse are located at (-3 + 9.72, -1) and (-3 - 9.72, -1), which simplifies to (-11, -1) and (4.72, -1).

7. The vertices of the ellipse can be found by applying the distance between the center and the semi-major axis. Since the vertex is the endpoint of the major axis, the distance from the center to the vertex is "a".

Therefore, the vertex coordinates are (-3 + sqrt(211), -1) and (-3 - sqrt(211), -1), which simplifies to (5, -1) and (-11, -1).

Finally, to graph the equation, plot the coordinates of the center, foci, and vertices on a coordinate plane. Draw an ellipse with the center (-3, -1), semi-major and semi-minor axes lengths calculated in steps 5 and 6, and a symmetric shape based on these points.