Question

An equation of an ellipse is given. 9x^2−36x+y^2 +2y+1=0 (a) Find the center, vertices, and foci of the ellipse. Center (x,y)=( focus (x,y)=()( smaller y-value) focus (x,y)= (larger y-value) vertex (x,y)= (smaller y-value) vertex (x,y)=( (larger y-value) (b) Determine the lengths of the major and minor axes. Major axis units minor axis units c) Sketch a praph of the ellitse

Answer 1

(a)
To find the center of the ellipse, we need to complete the square for both the x and y terms.

9x^2 - 36x + y^2 + 2y + 1 = 0

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

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

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

So the center of the ellipse is (2, -1).

To find the vertices, we need to find the distance from the center to the endpoints of the major axis. Since the major axis is along the x-axis, we use the formula a^2 = 36/9 = 4 to find the distance from the center to the endpoints.

Vertex 1: (2 - 2, -1) = (0, -1)
Vertex 2: (2 + 2, -1) = (4, -1)

To find the foci, we use the formula c^2 = a^2 - b^2, where a = 2 and b is the distance from the center to the endpoints of the minor axis. Since the minor axis is along the y-axis, we use the formula b^2 = 36/1 = 36. So c^2 = 4 - 36 = -32, which is not a real number. Therefore, the ellipse does not have any foci.

(b)
The length of the major axis is the distance between the two vertices, which is 4 units. The length of the minor axis is the distance between the two endpoints of the minor axis, which is 2 times the square root of 9, or 6 units.

(c)
Here's a sketch of the ellipse:

```
|
-1 |
|
|
| ****
| ** **
| * *
| * *
| * *
| * *
| * *
| ** **
| ****
|
|
|
|
|
|
|
|
+-------------------------
2 4
```